Applied Mathematics 3 Problems And Solutions Pdf
This book includes sample examples and practice problems for advanced engineering mathematics (ordinary differential equation).
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ADVANCED ENGINEERING
MATHEMATICS
Table of Contents
EXERCISE 1 .................................................................................................................................................... 3
EXERCISE 2 .................................................................................................................................................. 24
EXERCISE 3 .................................................................................................................................................. 47
EXERCISE 4 .................................................................................................................................................. 63
EXERCISE 5 .................................................................................................................................................. 80
EXERCISE 6 .................................................................................................................................................. 92
EXERCISE 7 ..................................................................................................... Error! Bookmark not defined.
EXERCISE 8 ..................................................................................................... Error! Bookmark not defined.
2
PRACTICE PROBLEMS FOR ADVANCED ENGINEERING MATHEMATICS
EXERCISE 1
FIRST ORDER ODE
EXAMPLE 1.1
= is the general solution of an ODE = . Plot the solution curves.
The family of solution for different values of (= 0.1, 0.2,… .0.9, 1.0) is shown in Figure
1.1.
EXAMPLE 1.2
Determine the general solution of an ODE + = . Plot the solution curves.
Solution: Given differential equation can be rearranged as
or,
or,
or,
Figure 1.1: The family of
solutions for =
3
Integrating both sides,
( 1) = +
Simplifying,
Taking anti-log of both sides, general solution can be expressed as
The family of solution for different values of (= 10, 8, … .8, 10) is shown in Figure 1.2.
Figure 1.2: The family of solutions for 1 =
EXAMPLE 1.3
Solve the following boundary value problems (BVP).
(+ )
= , ( )=
Solution: Given differential equation can be rearranged as
or,
4
or,
Integrating both sides,
Substituting given condition (3 )= ,
Hence, the solution can be expressed as
EXAMPLE 1.4
Determine the general solution of an ODE = + .
Solution: Dividing both sides of the given differential equation by
[a]
Assuming that = , differentiating w. r. t.
[b]
Substituting Equation [b] into Equation [a],
or,
or,
Integrating both sides,
Substituting = / , the general solution can be expressed as
5
EXAMPLE 1.5
Determine the general solution of an ODE:
=
+
Solution: Given differential equation can be rearranged as
[a]
Assuming that = + , differentiating w. r. t.
[b]
Substituting Equation [b] into Equation [a],
or,
or,
Integrating both sides,
Substituting = + , the general solution can be expressed as
EXAMPLE 1.6
Solve the following boundary value problems (BVP).
=( ) + , ( )=
6
Solution: Dividing both sides of the given differential equation by
[a]
Assuming that = , differentiating w. r. t.
[b]
Substituting Equation [b] into Equation [a],
or,
or,
Integrating both sides,
Substituting = / , the general solution can be expressed as
Substituting given condition ( 1) = 0 ,
Hence, the solution can be expressed as
EXAMPLE 1.7
Determine the general solution of an ODE:
7
= + +
+
Solution: Substituting = + , = + in the given differential equation,
= 2 + 5 + + 5+ 1
5 + 2 + + 2 1
[a]
Equation [a] can be converted into homogeneous form when
[b]
Solving Equations [b] , we get
Then homogeneous form of Equation [a] can be expressed as
[c]
Assuming that = , differentiating w. r. t.
[d]
Substituting Equation [d] into Equation [c],
or,
or,
=2 + 5 5 2
5 + 2
or,
or,
or,
7
2(1 )+ 3
2(1 + ) = 2
8
Integrating both sides,
7
2 ( 1 ) + 3
2 ( 1 + ) = 2 + 2 1
3 ( 1 + ) 7 ( 1 )= 4 ( 1)
or,
(1 + )
(1 ) = (1 )
or,
Substituting = / ,
Substituting = 1/3 and = + 1/3 the general solution can be expressed as
EXAMPLE 1.8
Determine the general solution of an ODE:
= + +
+ +
Solution: Given differential equation can be rearranged as
= 2 + 3 + 4
2(2 + 3 )+ 5
[a]
Assuming that = 2 + 3 , differentiating w. r. t.
∴
[b]
Substituting Equation [b] into Equation [a],
9
or,
or,
or,
= 3 +12 + 4 +10
2 + 5 = 7+22
2 + 5
or,
or,
or,
Integrating both sides,
Substituting = 2 + 3 , the general solution can be expressed as
14( 2 + 3 ) 9 ( 14 +21 +22) =49 +
EXAMPLE 1.9
Determine the general solution of an ODE:
+ +
=
Solution: We first check the exactness of the equation. For this comparing given ODE
with the standard form of an exact ODE, we get
[a]
[b]
Differentiating w. r. t. and differentiating w. r. t.
10
[c]
[d]
Here / =/ , hence given ODE is an exact ODE.
The general solution can be determined as
=
+ 2 2 + ( )
[e]
Again, we know
or,
or,
Then substituting ( ) into [e], the general solution can be expressed as
Alternatively, the general solution can also be determined as
11
Again, we know
or,
or,
Then substituting ( ) into [f], the general solution can be expressed as
EXAMPLE 1.10
Solve the initial value problem
+( ) = , ()=
Solution: We first check the exactness of the equation. For this comparing given ODE with the
standard form of an exact ODE, we get
Differentiating w. r. t. and Differentiating w. r. t.
[c]
[d]
Here / =/ , hence given ODE is an exact ODE.
The general solution can be determined as
12
Again, we know
or,
or,
Then substituting ( ) into [e], the general solution can be expressed as
Substituting given condition ( 5) = 0 ,
Hence, the solution can be expressed as
EXAMPLE 1.11
Determine the particular solution of an ODE subject to given initial condition
( + ) = , ( )=
Solution: Given ODE is not exact, hence
Then the integrating factor is determined as
=
=
= = =
13
The given ODE becomes an exact ODE when it is multiplied by the integrating factor
thus obtained. Then
Then the general solution can be determined as
Again, we know
or,
or,
∴
Then substituting ( ), the general solution can be expressed as
Substituting the given initial condition, we get
∴
Then substituting , we get the particular solution as
14
∴
EXAMPLE 1.12
Determine the general solution of the following ODE
(+ )
+ =
Solution: Given ODE can be rearranged b y dividing it with ( 1 + )
Comparing with the standard form of first order linear ODE [41] , we get
Then,
= = 2
1 + = ( 1 + )
∴
and
Then the general solution of the given ODE is given as
=1
1 + ( 1 + ) 6
1 + + 1
1 +
Then the general solution can also be expressed as
15
EXAMPLE 1.13
Determine the particular solution of an ODE subject to given initial condition
+ =
, ( / ) =
Solution: Given ODE can be rearranged by dividing it with
Comparing with the standard form of first order linear ODE [41] , we get
Then,
∴
and
Then the general solution of the given ODE is given as
Substituting the given initial condition, we get
∴
Then substituting , we get the particular solution as
16
∴
EXAMPLE 1.14
Determine the general solution of the following ODE
= ( + )
Solution: Given ODE can be rearranged as
Comparing with the standard form of Bernoulli Equation [49] , we get
Then,
Then the general solution as a dependent variable of the given ODE is given as
Substituting = = = Then the general solution can also be expressed as
17
EXERCISE
Solve all problems analytically and verify using computer application.
A: Separable ODEs
1. Find a general solution for the following differential equations. Also plot the solution curves.
(a)
(b)
(c)
(d)
dy
dx = xy +2y x 3
xy 3y + x 3
(e)
dy
dx = sin ( x + y ) + sin(x y)
(f)
(g)
(h)
(i)
(j)
(1 e ) sec ydy
dx = e tan y
2. Solve the following boundary value problems (BVP).
(a)
(b)
(c)
(d)
2y dy
dx = x ( 12y) , y( 1) = 1
(e)
(f)
dy
dx +3y( y + 1) sin 2x = 0, y( 0) = 1
(g)
(h)
cos x ( e y)dy
dx = e sin 2x , y( 0) = 0
(i)
y dy
dx = sin x
1 x , y( 0) = 0
(j)
ln y dy
dx = 3x y, y ( 2 ) = e
3. Determine the value of , given that
=
and that y = 0 when x = 2 and when x = 6.
4. Find an explicit solution of the given initial-value problem. Use a graphing utility to plot the
graph of each solution. Compare each solution curve in a neighborhood of (0, 1).
18
=( 1) + 0.01, ( 0) = 1
=( 1) 0.01, ( 0) = 1
Compare and comment upon the solutions of each case.
B: Reducible to Separable Form
5. Find a general solution for the following differential equations. Also plot the solution curves.
/ 1
+ 1 + /
= 0
6. Solve the following boundary value problems (BVP).
= + 3
, ( 1) = 0
( 1)
+ = 0, ( 1)=
/
+ / = 0, ( 1) = 0
=( 2 + ) + 7, ( 0) = 0
= 3 + 2
3 + 2 + 2 , ( 1 ) = 1
C: Homogeneous and Nonhomogeneous ODE
7. Find a general solution for the following differential equations. Also plot the solution curves.
19
D: Exact ODE
8. Find a general solution for the following differential equations. Also plot the solution curves.
( 2 ) + (3 + ) = 0
3
2 + +
+ + 6 = 0
( + ) + ( + 1 ) = 0
[2 (+ ) 2 (+ )] 2 (+ ) = 0
(4 + ) ( + 1 3 ) = 0
1 + 3 / + 3 / 1
= 0
( 2 2 2 + 2 ) + ( 2 3 ) = 0
+ 2 ( ) +[ + 2 ( )] = 0
9. Solve the following boundary/initial value problems (BVP/IVP).
2 +( 1 + ) = 0, ( 0) = 1
2 + ( 3 ) = 0, ( 1 ) = 1
( 3 ) + ( 3 ) = 0, (0 )= 1
(+ ) + ( 2 ) = 0, (2 )=
( + ) +2 1
1 + = 0, ( 0 ) = 1
( + ) + = 0, (0 )= 1
[ ( 2 ) ( 2 )] 2 ( 2 ) = 0, ( /12 )= /8
20
( + + 1) + 3 ( 6) = 0, ( 0) = 1
1 + /
/ +/ = 0, ( 1) = 5
( 3 2 ) + (2 + ), (0 )=
10. Find the value of for which the given equation is exact and then solve it using that value of
.
( + ) + (+ ) = 0
E: Reducible to Exact ODE
11. Find a general solution for the following differential equations.
4
+3
+3
+ 4 = 0
( + ) + (5 + ) = 0
(2 + )(1 + ) = 0
12. Verify that the given differential equation is not exact. Multiply the given differential
equation by the indicated integrating factor ( , ) and verify that the new equation is exact.
Then solve.
2 + + 2
= 0
13. Show that the given ODE is not exact and that an integrating factor depending upon alone
or alone does not exist. If possible, find an integrating factor in the form ( , )= ,
21
where and are suitably chosen constants. If such ( , ) can be found, then use it to
obtain the general solution of the differential equation; if not, state that.
(3 2 ) + (2 2 ) = 0
(3 + 2 ) + (3 + 4 ) = 0
14. Solve the following boundary/initial value problems (BVP/IVP).
+( + 4 ) = 0, ( 4) = 0
( + 5 ) (+ ) = 0, (0 )= 1
2( 2) + 3 = 0, ( 3) = 1
( + ) + ( 1 ), (0 )= 1
F: Linear ODE
15. Find a general solution for the following differential equations. Also plot the solution curves.
(1 + )
+( + ) = 0
16. Solve the following boundary/initial value problems (BVP/IVP).
(+ 2 )
= 0, (0 )= 3
+ 1 = 2 , ( 1) = 3
+ 2
=
, ( )= 0
+ = , ( 0) = 2.5
22
(+ 1 )
+ = , ( 1) =10
( )
+ = 0, ( 1)=
6
G: Bernoulli equation
17. Find a general solution for the following differential equations. Also plot the solution curves.
18. Solve the following boundary/initial value problems (BVP/IVP).
2
= ( ) , ( 0) = 1
23
PRACTICE PROBLEMS FOR ADVANCED ENGINEERING MATHEMATICS
EXERCISE 2
SECOND ORDER ODE
EXAMPLE 2.1
The functions = and = are solutions of the homogeneous linear ODE
+ =
Verify any linear combination of and is also the solution of the given ODE.
Solution: Assume that = + , a linear combination of and , is also a solution of
the given ODE. Again, assuming = 2 and = 3
Differentiating twice, we get
Substituting [a] and [b] into the given ODE,
= 8 2 + 12 2 + 4( 2 2 3 2 )
= 8 2 + 12 2 + 8 2 12 2 = 0
It verifies the superposition principle for the given homogeneous linear ODE.
EXAMPLE 2.2
Determine the general solution of an ODE = .
Solution: Assuming = as the solution of the given ODE, we get the characteristic
equation
Then the general solution is given by
24
EXAMPLE 2.3
Solve the following initial value problem
+ + = , ()= , ()=
Solution: Assuming = as the solution of the given ODE, we get the characteristic
equation
Then the general solution is given by
Differentiating [a] ,
Substituting given initial condition (0 )= 4 into Equations [a],
Again substituting given initial condition (0) = 6 and the value of into Equations [b],
Substituting and into Equations [a], we get the particular solution as
EXAMPLE 2.4
Solve the following initial value problem
+ + = , ()= , ()=
Solution: Assuming = as the solution of the given ODE, we get the characteristic
equation
or,
Then the general solution is given by
25
Differentiating [a] ,
= 2 ( + )+ ( + )
Substituting given initial condition (0 )= 2 into Equations [a],
Again substituting given initial condition (0) = 5 and the value of into Equations
[b],
Substituting and into Equations [a], we get the particular solution as
EXAMPLE 2.5
Solve the following initial value problem
+ = , ()= , ()=
Solution: Assuming = as the solution of the given ODE, we get the characteristic equation
Then the general solution is given by
Differentiating [a] ,
Substituting given initial condition (1 )= 1 into Equations [a],
Again substituting given initial condition (1) = 0 into Equation [b],
Solving simultaneous Equations [c] and [d],
26
Substituting and into Equations [a], we get the particular solution as
EXAMPLE 2.6
Solve the following initial value problem
+ + = , ()= , ()=
Solution: Assuming = as the solution of the given ODE, we get the characteristic equation
Then the general solution is given by
=( + ) = +
[a]
Differentiating [a] ,
[b]
Substituting given initial condition (1 )= 4 into Equation [a],
Again substituting given initial condition (1 )= 2 into Equation [b],
Solving simultaneous Equations [c] and [d],
Substituting and into Equations [a], we get the particular solution as
EXAMPLE 2.7
Determine the general solution of an ODE:
27
+ =
Solution: Assuming = as the solution of the given ODE, we get the characteristic equation
Then the general solution is given by
= [ ( )+ ( )]
EXAMPLE 2.8
Find the Wronskian for the given basis of solutions. Also show linear independence by
using quotients.
,
Solution: Here = and = .
which is not a constant number. Hence
and
are linearly independent.
Differentiating and w. r. t. ,
=2 , = ( 1 2 )
It can also be verified by Wronskian as
= . ( 1 2 ) .( 2 )
Wronskian can also be determined by using quotients
=
=
( ) = 1. = 0
28
=
=
( ) = 1
. = 0
EXAMPLE 2.9
For the given basis of solutions: (a) Find a second-order homogeneous linear ODE for
which the given functions are solutions. (b) Show linear independence by the Wronskian.
(c) Solve the initial value problem.
, , ( ) = , ( ) =
Solution: Here = and = . Given basis is the solution for Euler– Cauchy equation
having real repeated root with = 2. Then its characteristic equation is given by
Comparing it with the standard form of Euler–Cauchy equation we get coefficients as
Hence the ODE havine gicven functions as solutions can be expressed as
3 + 4 = 0
Differentiating and w. r. t. ,
Wronskian is the given as
= .( 2 + )
.( 2 )
Hence and are linearly independent
Then the general solution is given by
Differentiating [a] ,
Substituting given initial condition (1 )= 4 into Equation [a],
29
Again substituting given initial condition ( 1) = 6 and the value of into Equation [b],
Solving simultaneous Equations [c] and [d],
Substituting and into Equations [a], we get the particular solution as
EXAMPLE 2.10
Find a (real) general solution.
+ + =
Solution: Assuming = as the solution of the homogeneous part of the given ODE, we get
the characteristic equation
Then the general solution of homogeneous form of given ODE is
For given forcing function ( ) , we assume the particular solution as
Differentiating successively,
Since is also the solution, it should satisfy the given ODE, i.e.,
4 + 3( 2 ) + 2( ) = 24
Substituting into Equation [a],
30
Then the complete general solution of the given ODE is
= + = + + 2
EXAMPLE 2.11
Find a (real) general solution.
+ + =
Solution: Assuming = as the solution of the homogeneous part of the given ODE, we get
the characteristic equation
or,
Then the general solution of homogeneous form of given ODE is
For given forcing function ( ) , we assume the particular solution as
[a]
Differentiating successively,
Since is also the solution, it should satisfy the given ODE, i.e.,
( )+ 2 ( + )+ 5 ( + )=20
( + 2 + 5 ) + ( 2 + 5 ) =20
(4 + 2 ) + (2 + 4 ) =20
Comparing coefficients of and
Solving simultaneous Equations [b] and [c] for and , we get
31
Substituting and into Equation [a],
Then the complete general solution of the given ODE is
= + = ( 2 + 2 )+ ( 2 + 4 )
EXAMPLE 2.12
Find a (real) general solution.
++ = +
Solution: Assuming = as the solution of the homogeneous part of the given ODE, we get
the characteristic equation
Then the general solution of homogeneous form of given ODE is
For given forcing function ( ) , we assume the particular solution as
[a]
Differentiating successively,
Since is also the solution, it should satisfy the given ODE, i.e.,
( )+ 2 ( + )+ ( + + )
= 4 + 2
4 +( + 2 + ) +( 2 + ) = 4 + 2
4 +( 2 ) +( 2 ) = 4 + 2
Comparing coefficients of , and
32
Substituting , and into Equation [a] ,
Then the complete general solution of the given ODE is
= + =( + ) +
EXAMPLE 2.13
Find a (real) general solution.
=
Solution: Assuming = as the solution of the homogeneous part of the given ODE, we get
the characteristic equation
Then the general solution of homogeneous form of given ODE is
For given forcing function ( ) , we assume the particular solution as
Differentiating successively,
Since is also the solution, it should satisfy the given ODE, i.e.,
(4 + 4 ) (2 + ) 2 = 3
Substituting into Equation [a],
Then the complete general solution of the given ODE is
33
EXAMPLE 2.14
Solve the initial value problem.
+ + = , ( )= , ()=
Solution: Assuming = as the solution of the homogeneous part of the given ODE, we get
the characteristic equation
Then the general solution of homogeneous form of given ODE is
For given forcing function ( ) , we assume the particular solution as
[a]
Differentiating successively,
Since is also the solution, it should satisfy the given ODE, i.e.,
2 + 4( 2 + ) + 4( + + ) = 8
4 +( 8 + 4 ) +( 2 + 4+ 4 ) = 8
Comparing coefficients of
Comparing coefficients of and then substituting value of
Comparing constant term and then substituting values of and
Substituting , and into Equation [a] ,
34
Then the complete general solution of the given ODE is
= + =( + ) +(2 4 + 3)
Differentiating [b] ,
=( 2 2 ) +(4 4)
Substituting given initial condition (0 )= 0 into Equations [b],
Again substituting given initial condition (0 )= 1 and the value of into Equations
[c],
Substituting and into Equations [b] , we get the complete particular solution as
EXAMPLE 2.15
Solve the initial value problem.
= , ( ) = , ( ) =
Solution: Assuming = as the solution of the homogeneous part of the given ODE, we get
the characteristic equation
Then the general solution of homogeneous form of given ODE is
For given forcing function ( ) , we assume the particular solution as
Differentiating successively,
35
Since is also the solution, it should satisfy the given ODE, i.e.,
( ) 2 ( + ) 3 ( + )
= 5
( 2 3 ) + ( + 2 3 ) = 5
(4 2 ) + (2 4 ) = 5
Comparing coefficients of and
Solving simultaneous Equations [b] and [c] for and , we get
Substituting and into Equation [a],
Then the complete general solution of the given ODE is
= + = + + 1
2 +
[d]
Differentiating [d] ,
= + 2 +1
2 +
[e]
Substituting given initial condition (0 )= 0 into Equation [d],
[f]
Again substituting given initial condition ( 0) = 1 into Equation [e],
Solving simultaneous Equations [f] and [g] for and , we get
Substituting and into Equations [d] , we get the complete particular solution as
36
= 3
2 2 + 1
2 +
EXAMPLE 2.16
Solve the initial value problem.
+ = , ()= , ()=
Solution: Assuming = as the solution of the homogeneous part of the given ODE, we get
the characteristic equation
Then the general solution of homogeneous form of given ODE is
For given forcing function ( ) , we assume the particular solution as
[a]
Differentiating successively,
=(+ 3 ) 3 + (3 + )3
=(9 + 6 ) 3 + (6 9 )3
Since is also the solution, it should satisfy the given ODE, i.e.,
(9 + 6 ) 3 + (6 9 ) + 9 3 + 9 3 = 6 3
(9 + 6 + 9 ) 3 + (6 9 + 9 ) 3 = 6 3
6 3 +( 6 ) 3 = 6 3
Comparing coefficients of 3 and 3 , we get
Substituting and into Equation [a],
Then the complete general solution of the given ODE is
37
= + = 3 + 3 + 3
[b]
Differentiating [b] ,
= 3 3 + 3 3 + 3 3 3
Substituting given initial condition (0 )= 1 into Equation [b],
Again substituting given initial condition (0 )= 1 into Equation [c],
Substituting and into Equations [b] , we get the complete particular solution as
= 3 1
3 3+ 3
EXAMPLE 2.17
Find a (real) general solution.
++ =
Solution: Assuming = as the solution of the homogeneous part of the given ODE, we get
the characteristic equation
or,
Then the basis of homogeneous form of given ODE and corresponding Wornskain are
given as
= ( ) ( )
= ( + + )
Then the general solution of homogeneous form of given ODE is
38
Then its particular solution can be determined as
= ( 4 )
+ ( 4 )
= 4 + 4
= 4 1
2 + 4
= 2 + 4
Then the complete general solution of the given ODE is
= + = ( + )+ ( 2 + 4 )
EXERCISE
Solve all problems analytically and verify using computer application.
A: Homogeneous Linear ODEs of Second Order
1. Check which of the following ODEs follow superposition principle with the given
fundamental basis. Comment why it is satisfied or why not.
= 1 + , = 1 +
2. Reduce to first order and solve.
39
(i)
(j)
3. The indicated function () is a solution of the given equation. Use reduction of order or
direct formula to find a second solution ( ) .
9 12 + 4 = 0, =/
+ 2 = 0, = cos ( )
(1 ) + 2 = 0, = 1
(i)
2
1 + + 2
1 + = 0, =
(j)
+1
+1 1
4 = 0, = 1
4. Verify that the given functions are linearly independent and form a basis of solutions of the
given ODE. Solve the IVP. Plot the solution.
4 +25 = 0, ( 0) = 3.0, ( 0) = 2.5
3 4 = 0, ( 0) = 1, ( 0) = 2,
+ 0.6 + 0.09 = 0, ( 0) = 2.2, ( 0) = 0.14,
+ 2 + 2 = 0, ( 0) = 0, ( 0) =15,
3 = 0, (1 )= 3, (1 )= 0,
+ = 0, (1 )= 4.3, (1 )= 0.5,
B: Homogeneous Linear ODEs with Constant Coefficients
5. Find a general solution for the following differential equations.
40
6. Solve the following initial value problems (IVP).
= 0, ( 0) = 2, (0 )= 2
+ 3 = 0, (0 )= 2, (0 )= 3
+ 2 3 = 0, (0 )= 6, (0 )= 2
4 5 = 0, (1 )= 0, (0 )= 5
+ 8 9 = 0, (1 )= 1, (1 )= 0
4 4 3 = 0, ( 0) = 1, ( 0) = 5
8 2 = 0, ( 0) = 0.2, ( 0) = 0.325
+ 4 + 4 = 0, (0 )= 1, (0 )= 1
9 24 +16 = 0, ( 0) = 3, ( 0) = 3
+ = 0, ( /3) = 2, ( /3) = 4
+ + 2 = 0, (0 )= 0, (0 )= 0
2 + 3 = 0, (0 )= 4, (0 )= 1
4 + 5 = 0, (0 )= 2, (0 )= 5
4 +53 = 0, ( )= 3, ( /2) = 2
+ 4 + 5 = 0, (0 )= 2, (0 )= (0)
7. Solve the following boundary value problems (BVP).
+ 4 + 3 = 0, (0 )= 1, ( 1) = 0
10 +25 = 0, (0 )= 1, (1 )= 0
+ 4 = 0, (0 )= 0, ( )= 0
2 + 2 = 0, ( 0) = 1, ( ) = 1
+ 2 +17 = 0, ( 0) = 1, ( /4) = 0
+ 2 +26 = 0, ( 0) = 1, ( /4) = 0
8. Find an ODI of the form + + = 0 for the given basis.
(g)
(h)
. 2.1 , . 2.1
41
C: Euler–Cauchy Equations
9. Find a general solution for the following differential equations.
10. Solve the following initial value problems (IVP).
= 0, ( 2) = 5, ( 2) = 8
4 = 0, ( 1) = 0, ( 1) = 3
+ = 0, ( 2) = 1, (2 )= 3
+ 9 = 0, ( 2) = 1, (2 )= 2
4 + 6 = 0, ( 1) = 0.4, (1 )= 0
+ 5 21 = 0, (2 )= 1, (2 )= 0
+ 3 + 0.75 = 0, ( 1) = 1, (1 )= 1.5
3 + 4 = 0, ( 1) = 5, (1 )= 3
+25 +144 = 0, ( 1) = 4, (1 )= 0
+ + = 0, ( 1) = 1, (1 )= 0
+ 3 + 2 = 0, ( 1) = 0, (1 )= 2
+ 7 +13 = 0, ( 1) = 1, ( 1) = 3
4 + 5 = 0, ( 1) = 0, ( 1) = 1
11. Solve the following boundary value problems (BVP).
7 +12 = 0, (0 )= 0, (1 )= 0
3 + 5 = 0, ( 1) = 0, ( ) = 1
12. Find an ODI of the form + + = 0 for the given basis.
42
(e)
/ 1
2, / 1
2
D: Existence and Uniqueness of Solutions: Wronskian
13. Find the Wronskian for the given basis of solutions. Also show linear independence by using
quotients.
(g)
(h)
14. For the given basis of solutions: (a) Find a second-order homogeneous linear ODE for which
the given functions are solutions. (b) Show linear independence by the Wronskian. (c) Solve
the initial value problem.
1, ; ( 0) = 1, ( 0) = 1
, ; (0 )= 1, (0 )= 4
3 , 3 ; ( /3) = 0, ( /3) = 1
7 , 7 ; (0 )= 2, (0 )= 0
2 , 2 ; ( 0) = 1, ( 0) = 0
, ; (1 )= 2, (1 )= 4
E: Nonhomogeneous ODEs; Method of Undetermined Coefficients
15. Find a general solution for the following differential equations.
(c)
(d)
6 +25 =50 36 63 +18
43
(k)
(l)
9 +14 = 3 5 2 + 7
+ 3 + 2 = ( + 1) 2 + 3 + 4
16. Solve the following initial value problems (IVP).
3 + 4 = 8, ( 0) = 0, ( 0) = 0
+ 5 + 4 =20 , (0 )= 0, (0 )= 2
+ 4 + 3 = , ( 0) = 2, (0 )= 1
(d)
2 + = 3 , (0 )= 2
3, ( 0 ) = 13
3
+ 4 = 2, ( /8) = 1/2, ( /8) = 2
+ 2 = 2 , (0 )= 0, (0 )= 1
(g)
9 12 + 4 = 3 1, ( 0) = 0, ( 0) = 4
3
2 + 3 + 2 = 1 9 , ( 0) = 0, ( 0) = 1
3 4 = 3 , (0 )= 0, (0 )= 0
+ 6 + 9 = , (0 )= 1, (0 )= 6
+ 4 = 16 2 , (0 )= 0, (0 )= 0
+ 2 +10 = 17 37 3 , (0 )= 6.6, (0 )= 2.2
4 5 = , (0 )= 0, (0 )= 2
+ 6 +13 = , (0 )= 2, (0 )= 1
+ 4 + 4 = ( 3 + ) , ( 0) = 2, (0 )= 5
+ 4 + 5 = 1 + , (0 )= 2, (0 )= 1
+ 4 + 5 = + , (0 )= 1, (0 )= 0
17. Solve the following boundary value problems (BVP).
+ = + 1, (0 )= 5, (1 )= 0
2 + 2 = 2 2, ( 0) = 5, ( )=
+ 3 = 6 , (0 )= 0, (1 )+ ( 1) = 0
+ 3 = 6 , (0 )+ (0) = 0, (1 )= 0
44
F: Nonhomogeneous ODEs; Solution by Variation of Parameters
18. Find a general solution for the following nonhomogeneous differential equations.
(g)
(h)
(i)
(j)
3 6 +30 =15 +
19. Find a general solution for the following nonhomogeneous differential equations.
(c)
(d)
+ + 4 = 2 ( )
20. Solve the following initial value problems (IVP).
+ 2 8 = 2 + , ( 0) = 1, (0 )= 0
+ 4 + 5 = , (0 )= 0, (0 )= 0
2 + = + 1, ( 0) = 1, ( 0) = 0
+ = , (0 )= 4, (0 )= 3
+ 4 = 2 , (0 )= 0, (0 )= 0
+ 4 = 2 , ( /8) = 0, ( /8) = 0
+ 4 = 1 + 2 , (0 )= 0, (0 )= 0
+ = 2 , (0) = 0, (0 )= 0
4 = / , ( 0) = 1, ( 0) = 0
+ 5 + 6 = , ( 1) = 0, (1 )= 0
45
(k)
2 + =
, (1) = 0, (1 )= 1
4 + 4 =( 12 6 ) , (0 )= 1, (0 )= 0
21. Solve the following initial value problems (IVP).
(a)
+ = , ( 1) = 1, ( 1) = 1
2
6 = 8 , (1 )= 1, (1 )= 0
(c)
5 + 8 = 8 , 1
2 = 0, 1
2 = 0
4 + 6 = , (2 )= 2, (2 )= 7
+ 7 + 9 =27 , (1 )= 1, (1 )= 4
3 + 3 = 2 , (1 )= 0, (1 )= 0
2 + 2 =10 ( ) , (1 )= 3, (1 )= 0
46
PRACTICE PROBLEMS FOR ADVANCED ENGINEERING MATHEMATICS
EXERCISE 3
HIGHER ORDER ODE
EXAMPLE 3.1
Are the given functions linearly independent or dependent on any interval? Give reason.
( ) ,( + ) ,
Solution: Given functions =( 1) , =( + 1) and = will be linearly dependent
if they satisfy
+ + = 0 (for some k1 , k2 and k3 not all zero)
( 1) + ( + 1) + = 0
( + ) + (2 + 2 + )+ ( + )= 0
Equating coefficients of , and constant term we get
Substituting = from [a] or [c] into [b] , we get
Assuming = 1, we get
Hence the given functions are related as
Which can also be verified by evaluating Wronskian,
=
= ( 1) ( + 1)
2( 1)2( + 1)1
2 2 0 = 0
which also proves the linear dependency.
EXAMPLE 3.2
47
Show that the given functions are solutions and form a basis on any interval. Use
Wronskians.
, , , ; + =
Solution: If the given functions = 1 , = , = 3 and =3 form the basis of
the given ODE. Then its general solution is given by
= + + + = + + 3 + 3
Differentiating successively, we get
= 3 3 + 3 3
= 9 3 9 3
=27 3 27 3
=81 3 +81 3
Substituting and into given ODE,
81 3 +81 3 + 9( 9 3 9 3 ) = 0
or, 0 = 0
which proves that that the given functions are solutions and form a basis.
To verify linear independency, Wronskian can be evaluated,
=
= 1 3 3
0 1 3333
00 9 3 9 3
0 0 27 3 27 3
=243( 3 + 3 ) = 243 0
which also proves the linear independency.
EXAMPLE 3.3
Determine the general solution of an ODE + = .
Solution: Assuming = as the solution of the given ODE, we get the characteristic
equation
48
Then the general solution is given by
= + + +
EXAMPLE 3.4
Determine the general solution of an ODE + = .
Solution: Assuming = as the solution of the given ODE, we get the characteristic
equation
Then the general solution is given by
EXAMPLE 3.5
Solve the following initial value problem
+ + = , ()= , ( )= , ()=
Solution: Assuming = as the solution of the given ODE, we get the characteristic
equation
3 + 3 + 2 + 2 1 = 0
3 ( + 1)+ 2 ( + 1) ( + 1) = 0
Then the general solution is given by
[a]
Differentiating [a] successively,
49
= ( + ) +1
3
[b]
= 2 ( + ) +1
9
[c]
Substituting given initial condition (0 )= 0 into Equations [a],
Similarly, substituting given initial condition ( 0) = 1 into Equations [b],
[e]
Again substituting given initial condition (0) = 1 into Equations [c],
[f]
Solving simultaneous equations [d], [e] and [f] for , and , we get
Substituting , and into Equations [a] , we get the particular solution as
EXAMPLE 3.6
Determine the general solution of an ODE + + =.
Solution: Assuming = as the solution of the given ODE, we get the characteristic
equation
Then the general solution is given by
= [( + ) 2 +( + )] 2
EXAMPLE 3.7
Find a (real) general solution.
+ =( + )
50
Solution: Assuming = as the solution of the homogeneous part of the given ODE, we get
the characteristic equation
Then the general solution of homogeneous form of given ODE is
Given forcing function ( ) can be expanded as
For given forcing function ( ) , we assume the particular solution as
Differentiating successively,
Since is also the solution, it should satisfy the given ODE, i.e.,
64 + 8 2( 16 + 4 ) 5(4 + 2 )+ 6( + + )
= + 6 + 9
(64 32 20 + 6 ) + (8 8 10 + 6 ) + 6 = + 6 + 9
(18 ) + (4 ) + 6 = + 6 + 9
Comparing coefficients of , and constant term, respectively we get
Substituting , and into Equation [a],
Then the complete general solution of the given ODE is
= + = + + +1
18 3
2 + 3
2
51
EXAMPLE 3.8
Find a (real) general solution.
+ =
+
Solution: Assuming = as the solution of the homogeneous part of the given ODE, we get
the characteristic equation
Then the three linearly independent solutions (basis) of homogeneous form of given ODE are
Then the general solution of homogeneous form of given ODE is
For given forcing function ( ) , we can determine the particular solution by using method of
variation of parameters.
( )= ( ) ()
( ) ( ) + ( ) ()
( ) ( ) + ( ) ()
( ) ( )
[a]
where
= 1 2 3
1 2 3
1 2 3
=0
0
1
= 0
0
1
= 0
0
1
Substituting , , and into Equation [a] ,
52
= 3
2 3
1 + + 2 2
2 3
1 + + 2
2 3
1 +
= 1
2
1 + 1
1 + + 2 1
2
1 +
=1
2 1
2 ( + 1 ) [ ( + 1 ) ( ) ] + 1
2 2 ( + 1)
Then the complete general solution of the given ODE is
= + = + + +1
2 1
2 ( + 1 ) [ ( + 1) ( )]
+1
2 ( + 1)
EXAMPLE 3.9
Find a (real) general solution.
+ + + = + +
Solution: Assuming = as the solution of the homogeneous part of the given ODE, we get
the characteristic equation
Then the general solution of homogeneous form of given ODE is
For given forcing function ( ) , we assume the particular solution as
= + + +
[a]
Differentiating successively,
= + + + +
= 2 + + 2 2
= + 3 + 3 3
Since is also the solution, it should satisfy the given ODE, i.e.,
53
( + 3 + 3 3 )
+( 2 + + 2
2 )
+( + + + + )
+( + + + ) = + + sin
4 + 2 +( 2 2 ) +( 2 2 ) = + + sin
Comparing coefficients of and
Similarly, comparing coefficients of and
Solving simultaneous equations [b] and [c], for and , we get
and
Substituting , , and into Equation [a] ,
=1
4 + 1
2 1
4 1
4
Then the complete general solution of the given ODE is
= + = + + +1
4 + 1
2 1
4 1
4
EXAMPLE 3.10
Find a (real) general solution.
+ = +
Solution: Assuming = as the solution of the homogeneous part of the given ODE, we get
the characteristic equation
Then the general solution of homogeneous form of given ODE is
54
= + + +
For given forcing function ( ) , we assume the particular solution as
Differentiating successively,
= 2 + +
= 2 + 4 +
= 6 + 6 + +
=12 + 8 + + +
Since is also the solution, it should satisfy the given ODE, i.e.,
12 + 8 + 2 + +
2 2 + 4 + 2
+ 2 + + = +
8 + 4 + 4 = +
Comparing coefficients of , and
Substituting , and into Equation [a],
Then the complete general solution of the given ODE is
= + = + + + +1
8 + 1
4
EXAMPLE 3.11
Find a (real) general solution.
+ + = +
Solution: Assuming = as the solution of the homogeneous part of the given ODE, we get
the characteristic equation
55
Then the general solution of homogeneous form of given ODE is
For given forcing function ( ) , we assume the particular solution as
[a]
Differentiating successively,
=8 2 + 8 2
Since is also the solution, it should satisfy the given ODE, i.e.,
[16 2 + 16 2 ] + 2[4 2 4 2 ]+ [+ 2 + 2 ]
=3 + 2
+ 9 2 + 9 2 = 3 + 2
Comparing coefficients of constant term, 2 and 2
Substituting , and into Equation [a],
Then the complete general solution of the given ODE is
= + =( + ) + ( + ) +3+1
9 2
EXAMPLE 3.12
56
Find a (real) general solution.
+ =
Solution: Assuming = as the solution of the homogeneous part of the given ODE, we get
the characteristic equation
Then the general solution of homogeneous form of given ODE is
For given forcing function ( ) , we assume the particular solution as
Differentiating successively,
Since is also the solution, it should satisfy the given ODE, i.e.,
Substituting into Equation [a],
Then the complete general solution of the given ODE is
= + = + + +1
4 2
EXAMPLE 3.13
Solve the initial value problem.
+ + + = , ()= , ()= , ()= ,
()=
Solution: Assuming = as the solution of the homogeneous part of the given ODE, we get
the characteristic equation
57
Then the general solution of homogeneous form of given ODE is
= + +( 2 + 2 )
For given forcing function ( ) , we assume the particular solution as
[a]
Differentiating successively,
= +
Since is also the solution, it should satisfy the given ODE, i.e.,
( + + ) + 2 ( + )
+( ) + 8( + )
12( + + )=12
20 +( 12 + 6 ) ( 6 12 ) = 12
Comparing coefficients of
Similarly, comparing coefficients of and
Solving simultaneous equations [b] and [c], for and , we get
and
Substituting , and into Equation [a],
=1
20 2
5 4
5
58
Then the complete general solution of the given ODE is
= + = + +( 2 + 2 ) + 1
20 2
5 4
5
[d]
Differentiating [d] successively,
= 3 +( 2 2 + 2 2 ) 1
20 + 2
5 4
5
[e]
= + 9 +( 4 2 4 2 )+1
20 + 2
5 + 4
5
[f]
and
= 27 +( 8 2 8 2 ) 1
20 2
5 + 4
5
[g]
Substituting given initial condition (0 )= 3 into Equation [d],
[h]
Substituting given initial condition (0 )= 0 into Equation [e],
[i]
Substituting given initial condition (0 )= 1 into Equation [f],
[j]
Similarly, substituting given initial condition (0 )= 2 into Equation [g],
[k]
Solving simultaneous equations [h], [i], [j] and [k] for , , and ,
Substituting , , and into Equations [d] , we get the complete particular solution as
= 81
40 + 73
520 + 77
65 2 49
130 2 + 1
20 2
5 4
5
59
EXERCISE
Solve all problems analytically and verify using computer application.
A: Linear ODEs of Higher Order
1. Are the given functions linearly independent or dependent on any interval? Give reason.
2. Show that the given functions are solutions and form a basis on any interval. Use
Wronskians.
, , ; + 2 2 = 0
, , ; 3 + 3 = 0
1, , , ; + 2 + = 0
1, , 3 , 3; + 9 = 0
, , , ; + 2 + = 0
, 3 , 3 ; + 3 + 9 13 = 0
, , , ; 6 +25 = 0
, , 1/ ; + 2 + 2 = 0
, , ( ) ; + = 0
B: Homogeneous Higher Order Linear ODEs
3. Find a general solution for the following differential equations.
7 + 6 +30 36 = 0
+ 2 + 4 + 2 + 3 = 0
3 + 3 3 + 2 = 0
60
+ 4 + 6 + 4 + = 0
+ 2 11 12 +36 = 0
+ 5 2 10 + + 5 = 0
8 +26 40 +25 = 0
4. Find a general solution for the following differential equations.
+ 6 + 3 3 + = 0
+ 3 2 + 2 = 0
+ 6 + 7 + = 0
3 + 7 8 = 0
+ 6 3 + 3 2 3 + 4 = 0
5. Solve the following initial value problems (IVP).
+ 2 = 0, (0 )= 1, (0 )= 0, (0 )= 0
+ 2 5 6 = 0, (0 )= 0, (0 )= 0, (0 )= 1
(c)
+ 7 + 6 = 0, ( 0) = 1, (0 )= 0, (0 )= 2,
(0 )= 1
(d)
5 4 = 0, (0 )= 3, (0 )= 5, (0 )=11,
(0 )= 23, (0 )=47
+ 2 + 4 = 0, (0 )= 0, (0 )= 1, (0 )= 0
+ 3 + 7 + 5 = 0, (0 )= 1, (0 )= 0, (0 )= 0
4 + 8 +41 +37 = 0, ( 0) = 9, ( 0) = 6.5, ( 0) = 39.75
+ 4 = 0, (0 )= 1, (0 )= 1, (0 )= 0
6 = 0, (0 )= 0, (0 )= 1, (0 )= 0, (0 )= 0
+12 +36 = 0, (0 )= 0, (0 )= 1, (0 )= 7
+ 3 4 = 0, (0 )= 0, (0 )= 0, (0 )= 6
(l)
+ 4 = 0, (0 )=1
2, ( 0 ) = 3
2, ( 0 ) = 5
2, ( 0 ) = 7
2
9 400 = 0, (0 )= 0, (0 )= 0, (0 )=41, (0 )= 0
6. Solve the following initial value problems (IVP).
= 0, ( 0) = 1, ( 0) = 0, ( 0) = 0
2 = 0, ( 1) = 5, ( 1) = 0, ( 1) = 0, ( 1) = 0
+ 6 6 = 0, ( 1) = 2, (1 )= 1, (0 )= 4
61
2 = 0, (1 )= 2, (1 )= 0, (1 )= 0
+ 2 2 = 0, ( 0) = 1, (0 )= 0, (0 )= 0
C: Nonhomogeneous Higher Order Linear ODEs
7. Find a general solution for the following differential equations.
4 + 4 = 5 +
+ 3 + 2 = + 4 + 8
+ + + = sin 2 + 3
+ 2 3 = + 3 + 4
2 4 + 8 = + 8
5 + 8 4 = + 2 + 3
+ 6 +12 + 8 =12
+ = +4+
+ 2 + 2 = 3 + 2 +
8. Solve the following initial value problems (IVP).
(a)
3 + 2 = + , (0 )= 1, (0 )= 1
4, ( 0 ) = 3
2
=, ( /2) = 2, ( /2) = 1, ( /2) = 1
+ 4 = , ( 0) = 0, (0 )= 0, (0 )= 1
+ 8 = 2 5 + 8 , ( 0) = 5, (0 )= 3, (0 )= 4
+ = , (0 )= 2, (0 )= 1, (0 )= 2
+ = , (0 )= 2, (0 )= 1, (0 )= 1
+ 3 + 3 + =30 , (0 )= 3, (0 )= 3, (0 )= 47
2 + = 2 24 +40 , (0 )= 1/2, (0 )= 5/2, (0 )= 9/2
+ 2 + = 3 + 4, (0 )= 0, (0 )= 0, (0 )= 1, (0 )= 1
+ 2 + = , (0 )= 2, (0 )= 0, (0 )= 1, (0 )= 1
62
PRACTICE PROBLEMS FOR ADVANCED ENGINEERING MATHEMATICS
EXERCISE 4
SYSTEM OF ODES
EXAMPLE 4.1
Find a general solution of the given ODE by converting it to a system and compare with
method discussed earlier.
+ =
Solution: Given ODE can be converted into a system of ODE by setting
Differentiating,
= = 6 11 + 6 = 6 11 + 6
which can be expressed in matrix form as
= 0 1 0
0 0 1
6 11 6
Then characteristic equation of the given system is given by
Then eigenvector corresponding to = 1 is determined by
1 0
0 1
6 11 6
()
= 0
0
0
63
or,
110
0 1 1
6 11 5
()
= 0
0
0
Similarly, eigenvector corresponding to = 2 and = 3 can be determined to be
and
Then the general solution for the converted system of ODEs is given by
=
() +
() +
()
= 1
1
1 +1
2
4 + 1
3
9
Then the general solution for the given ODE is given by
EXAMPLE 4.2
Find a real general solution of the following systems.
= + +
=
=
Solution: Coefficient matrix for the given system of ODE is
= 121
6 1 0
121
Then characteristic equation of the given system is given by
1 2 1
6 1 0
121 = 0
(1 )(1 + ) 6 (2 2 + 2 ) (1 + )= 0
64
Then eigenvector corresponding to = 0 is determined by
1 2 1
6 1 0
121
()
= 0
0
0
or,
121
6 1 0
121
()
=0
0
0
Similarly, eigenvector corresponding to = 3 and = 4 can be determined to be
and
Then the general solution for the given system of ODEs is given by
=
() +
() +
()
= 1
6
13 + 2
3
2 + 1
2
1
EXAMPLE 4.3
Find a real general solution of the following systems.
=
=
Solution: Coefficient matrix for the given system of ODE is
= 82
2 4
65
Then characteristic equation of the given system is given by
Then eigenvector corresponding to = 6 is determined by
8 2
2 4
( )
=0
0
or,
To determine linearly independent for repeated eigenvalue, we assume the solution of the form
() =
( ) +
Differentiating () and substituting into = , we get
( ) +
( ) +
=
() +
or,
( )
+
( ) +
=
() +
Since =
( )
+
=
or,
or,
66
() = 1
1 + 0
1
2
Then the general solution for the given system of ODEs is
= 1
1 + 1
1 + 0
1
2
EXAMPLE 4.4
Find a real general solution of the following systems.
= +
= +
= +
Solution: Coefficient matrix for the given system of ODE is
= 1 1 2
110
101
Then characteristic equation of the given system is given by
1 1 2
1 1 0
1 0 1 = 0
(1 )(1 ) (1 )+ 2 (1 )= 0
= 1, = 1 , = 1 +
Then eigenvector corresponding to = 1 is determined by
1 1 2
1 1 0
1 0 1
()
=0
0
0
or,
0 1 2
1 0 0
1 0 0
()
= 0
0
0
67
Similarly, the eigenvector corresponding to = 1 is determined by
1 1 2
1 1 0
1 0 1
()
=0
0
0
or,
1 2
1 0
1 0
()
= 0
0
0
Similarly, eigenvector corresponding to = 1 + can be determined to be
Then the general solution for the given system of ODEs is given by
= 0
2
1 + 1
( ) + 1
( )
EXAMPLE 4.5
Solve the following initial value problems
= + +
=
= + +
()= , ()= , ()=
Solution: Coefficient matrix for the given system of ODE is
= 1 1 4
0 2 0
1 1 1
68
Then characteristic equation of the given system is given by
1 1 4
0 2 0
1 1 1 = 0
Then eigenvector corresponding to = 1 is determined by
1 1 4
0 2 0
1 1 1
()
=0
0
0
or,
2 1 4
0 3 0
1 1 2
()
= 0
0
0
Similarly, eigenvector corresponding to = 2 and = 3 can be determined to be
and
Then the general solution for the given system of ODEs is given by
=
() +
() +
()
= 2
0
1 + 5
3
2 +2
0
1
[a]
Substituting given initial conditions ( 0) = 1, ( 0) = 3, ( 0) = 0 into Equation [a] , we get
2 5 2
0 3 0
1 2 1
= 1
3
0
Solving for , and , we get
69
Hence the particular solution for the given system of ODEs for the given initial conditions is
=1
2 2
0
1 5
3
2 + 5
2 2
0
1
EXAMPLE 4.6
Determine the general solution of a system of ODE
= + +
= +
Solution: Coefficient matrix for the given system of ODE is
= 2 2
23
Then characteristic equation of the given system is given by
Then eigenvector corresponding to = 1 is determined by
2 2
23
()
=0
0
or,
Similarly, eigenvector corresponding to = 2 can be determined to be
70
Then the general solution for the homogeneous part of the given system of ODEs is given by
()
=
( ) +
()
()
= 2
1 + 1
2
According the given forcing function g(t) we can assume the particular solution of the form
()
=
+
[a]
Differentiating y (p) , we get
() =
+
+
[b]
Since y (p) is also a solution, it should be satisfies on the given system of ode, i.e.,
+
+
= 2 2
23
+ 2 2
23
+ 1
1
or,
+
+
= 2 + 2
2 3 + 2 + 2
2 3 +1
1
or,
+
+
= 2 + 2
2 3 + 2 + 2
2 3 +1
1
or,
2 + 2
2 3 + 2 + 2 + 1
2 3 + 1 =0
0
+ 2
2 4 + + + 2 + 1
2 4 + 1=0
0
Comparing coefficients of t and constant term,
Assuming = 2 , we get the value of from Equation [c] or [d] as
Substituting = 2 and = 1, both Equations [e] and [f] reduce to
71
Assuming = 1 , we get the value of from Equation [g] as
Substituting , , and into Equation [a]
()
= 2
1 + 1
1
Then the complete general solution of the given system of ODE is given as
=
()
+
()
= 2
1 + 1
2 + 2
1 + 1
1
EXAMPLE 4.7
Determine the general solution of a system of ODE
= + /
= + /
Solution: Coefficient matrix for the given system of ODE is
= 2 2
8 6
Then characteristic equation of the given system is given by
Then eigenvector corresponding to = 2 is determined by
2 2
8 6
()
=0
0
or,
72
To determine linearly independent for repeated eigenvalue, we assume the solution of the form
() =
( ) +
Differentiating () and substituting into = , we get
( ) +
( ) +
=
() +
or,
( )
+
( ) +
=
() +
Since =
( )
+
=
or,
or,
Then the general solution for the homogeneous part of the given system of ODE is given by
()
=1
2 +1
2 +1
3
2
The fundamental matrix for the given system is given by
= +
2 2 + 3
2
73
Then its inverse is given by
= 4 3 2 +
4 2
Then,
= 4 3 2 +
4 2 /
3 / = 4 3
+ 6 +6
2
= 2 +3
2
Then the particular solution of the given system of ODE is given by
= +
2 2 + 3
2 2 +3
2
= +
2 2 + 3
2 + 3
2
= 1 + 3 2( + )
2 + 6 2 2 + 3
2
Then the complete general solution of the given system of ODE is given as
=1
2 + 1
2 + 1
3
2 + 1 + 3 2 ( + )
2 + 6 2 2 + 3
2
EXAMPLE 4.8
Solve the following initial value problem
=
= +
()= , ()=
74
Solution: Coefficient matrix for the given system of ODE is
= 0 1
4 0
Then characteristic equation of the given system is given by
Then eigenvector corresponding to = 2 is determined by
21
42
()
=0
0
Similarly, eigenvector corresponding to = 2 can be determined to be
Then the general solution for the homogeneous part of the given system of ODEs is given by
=
() +
( )
= 1
2 + 1
2
= ( + )2 + ( ) 2
2 ( ) 2 2( + ) 2
= 2+ 2
22 22
According the given forcing function g(t) we can assume the particular solution of the form
()
=
+
[a]
Differentiating y (p) , we get
() =
+
[b]
75
Since y (p) is also a solution, it should be satisfies on the given system of ode, i.e.,
+
= 0 1
4 0
+ 0 1
4 0
+ 5
17
or,
+
+ = + 5
4 4 +17
Comparing coefficients of and ,
Solving, we get
= = 0, = 1 and = 4
Substituting , , and into Equation [a]
()
= 4
0 + 0
1
Then the complete general solution of the given system of ODE is given as
=
()
+
()
= 2+ 2
22 22 + 4
0 + 0
1
Substituting given initial conditions (0 )= 5
Substituting given initial conditions (0 )= 1
Then the complete particular solution of the given system of ODE is given as
= 2 +1
2 2
2 2 2 +4
0 + 0
1
= 2+1
2 2 + 4
2 2 2 +
76
EXERCISE
Solve all problems analytically and verify using computer application.
A: Conversion of an nth-Order ODE to a System of ODEs
1. Find a general solution of the given ODEs by converting it to systems of ODEs and compare
with method discussed earlier.
B: Constant-Coefficient Homogeneous Systems
2. Find a real general solution of the following systems.
= 4 + 2
= 3 + 3
=4 + 2
=2.5 + 2
= 5 +
=2 + 3
= 4 + 5
=2 + 6
3. Solve the following initial value problems
77
=6
( 0) = 1, ( 0) = 19
= 4 +
( 0) = 1, ( 0) = 6
= 2 + 3
( 0) = 1, ( 0) = 5, ( 0) = 1
( 0) = 1, ( 0) = 2, ( 0) = 5
( 0) = 1, ( 0) = 7, ( 0) = 3
( 0) = 4, ( 0) = 6, ( 0) = 7
C: Nonhomogeneous Linear Systems of ODEs
4. Find a real general solution of the following systems.
= + 2 8
= + + 3
= 4 + 2 + 3
= 2 + +
= + 8 +
= +
78
= + +10
= 3 10
= + 3
= + + 3
= 2 + 2
= 4 + 2 2 2
= 2 +
= + 1
5. Solve the following initial value problems
79
PRACTICE PROBLEMS FOR ADVANCED ENGINEERING MATHEMATICS
EXERCISE 5
SERIES SOLUTION OF ODES
EXAMPLE 5.1
Find the radius of convergence of the following series.
(a)
(b)
(c)
Solution: For the series (a)
and
Then the radius of convergence R is given by
For the series (b)
Then the radius of convergence R is given by
For the series (c)
Then the radius of convergence R is given by
80
Convergence for all x (R =
∞
) is the best possible case, convergence in some finite interval the
usual, and convergence only at the center (R = 0) is useless.
EXAMPLE 5.2
Find the power series solution in powers of x.
+ =
Solution: We assume the solution of the form
= + + + + + ..
Differentiating , we get
= + 2 + 3 + 4 + ..
Substituting and into given differential equation, we get
( + 2 + 3 + 4 + )+ ( + + + + + )= 0
+( 2 + ) +( 3 + ) +( 4 + ) + = 0
Equating coefficients of every power of to zero, we get
Substituting , , and into Equation [a], we get the general solution of the given ODE
as
= 1 + ( /2) +( /2 )
2 . .
81
EXAMPLE 5.3
Find the power series solution in powers of x.
+ =
Solution: We assume the solution of the form
= + + + + + + ..
Differentiating successively, we get
= + 2 + 3 + 4 + 5 + ..
= 2 + 6 +12 +20 + ..
Substituting and into given differential equation, we get
(2 + 6 +12 +20 + )+ 4 ( + + + + )= 0
(2 + 4 )+ (6 + 4 )+ (12 + 4 ) + (20 + 4 ) + = 0
Equating coefficients of every power of to zero, we get
Substituting , , and into Equation [a], we get the general solution of the given ODE
as
= + 2 2
3 +2
3 +2
15 + ..
= 1 2 + 2
3 + 2
3 + 2
15
= 1 ( 2 )
2+ ( 2 )
4! . . +
2 2 ( 2 )
3! + ( 2 )
5! ..
EXAMPLE 5.3
82
Solve the initial value problem by a power series. Graph the partial sums of the powers up
to and including .
() = , ( ) =
Solution: We assume the solution of the form
= + + + + + + ..
Differentiating , we get
= + 2 + 3 + 4 + 5 + ..
Substituting and into given differential equation, we get
( 2 )( + 2 + 3 + 4 + 5 + )
( + + + + + + ) = 0
( + 2 + 3 + 4 + 5 + )
+( 2 4 6 8 10 …)
+( …) = 0
2 +( 4 ) +( 2 6 ) +( 3 8 )
+( 4 10 ) + = 0
Equating coefficients of every power of to zero, we get
=
8+ 3
8=
32
32 = 0
Substituting , , , and into Equation [a], we get the general solution of the given
ODE as
=
4
12 +
120 +..
[b]
Substituting given initial condition (0 )= 4 , we get
Substituting into Equation [b], we get the particular solution of the given ODE as
83
Plot of is shown below:
EXAMPLE 5.4
Find a basis of solutions by the Frobenius method.
+ + =
Solution: For Forbenius method, we assume solution of the form
Differentiating successively,
= ( + )( + 1)
Substituting , and into given differential equation, we get
2 ( + )( + 1)
+5 ( + )
+
= 0
84
[2 ( + )( + 1) + 5 ( + )]
+
= 0
In the first series we set = and = 2 in the second, thus = + 2 . Then
[2 ( + )( + 1) + 5( + )]
+
= 0
[a]
The lowest power is (take = 0 in the first series) and gives the indicial equation
and
Similarly taking = 1 in the first series, we get
[2 ( 1 + )( 1 + 1 ) + 5 ( 1 + )] = 0
[2 ( 1 + ) + 5 ( 1 + )] = 0
Since (2 + 7+ 5 ) 0 for both = 0 and =3/2, = 0.
Substituting = = 0 into Equation [a] , we get
[2 ( 1) + 5 ]
+
= 0
[b]
Equation [b] leads to the recurrence relation
Substituting = 2, 3, 4, …. successively, we get
=
44 =
14 × 44 =
616
85
Then the first solution is given by
=
14 +
616 = 1 1
14 + 1
616
Substituting = =3/2 into Equation [a] , we get
( 2 3)( 1)
2+ 5
2( 2 3)
+
= 0
[c]
Equation [b] leads to the recurrence relation
Substituting = 2, 3, 4, …. successively, we get
Then the second solution is given by
=
2 +
48 =
1 1
2 + 1
40
Then the general solution of the given ODE is given by
= 1 1
14 + 1
616 +
1 1
2 + 1
40
EXAMPLE 5.5
Find a basis of solutions by the Frobenius method.
+ + =
Solution: For Forbenius method, we assume solution of the form
86
Differentiating successively,
= ( + )( + 1)
Substituting , and into given differential equation, we get
( + )( + 1)
+ ( + )
+
= 0
[( + )( + 1) + ( + )]
+
= 0
In the first series we set = and = 1 in the second, thus = + 1 . Then
[( + )( + 1) +( + )]
+
= 0
[a]
The lowest power is (take = 0 in the first series) and gives the indicial equation
Substituting = = 0 into Equation [a] , we get
[ ( 1) + ]
+
= 0
[b]
Equation [b] leads to the recurrence relation
Substituting = 1, 2, 3, …. successively, we get
87
=
16 =
16 × 36 =
(4!)
Then the first solution is given by
= +
(2!)
(3!) = 1 +1
(2!) 1
(3!) +
Then second solution is given by method of reduction of order
where
=1
1 +1
4 2 1
36 3 +
=1
1 +1
4 2 1
36 3 +
= = + 2+5
4 2 + 22
27 3 +..
Then the general solution of the given ODE is given by
= 1 +1
(2!) 1
(3!) +
+ 1 + 1 2 +5
4 + 22
27 +. .
EXAMPLE 5.5
Find a basis of solutions by the Frobenius method.
88
( ) + =
Solution: For Forbenius method, we assume solution of the form
Differentiating successively,
= ( + )( + 1)
Substituting , and into given differential equation, we get
( ) (+ )(+ 1 )
+ [ ( + )( + 1) ( + )+ ]
In the first series we set = and = 1 in the second, thus = + 1 . Then
+ [( + 1)( + 2) ( + 1) + 1 ]
[a]
89
The lowest power is (take = 0 in the first series) and gives the indicial equation
Substituting = = 1 into Equation [a] , we get
( + 1)
+[ ( 1) + 1]
= 0
[b]
Equation [b] leads to the recurrence relation
( + 1) +[ ( 1) + 1] = 0
Substituting = 1, 2, 3, …. successively, we get
Then the first solution is given by
Then second solution is given by method of reduction of order
where
= = 1
1
= +1
= + 1
90
Then the general solution of the given ODE is given by
EXERCISE
19. Find a power series solution in powers of x .
20. Solve the initial value problem by a power series.
+ 3 + 2 = 0, ( 0) = 1, (0 )= 1
+ 3 + = 2 , ( 0) = 1, (0 )= 1
= + 1, (1 )= 1, (1 )= 0
(e)
2 + = 0, (1 )= 0, (1 )=1
2
21. Find a basis of solutions by the Frobenius method.
91
PRACTICE PROBLEMS FOR ADVANCED ENGINEERING MATHEMATICS
EXERCISE 6
Laplace Transform
EXAMPLE 6.1
Find the Laplace transform of the following functions.
(e)
Solution: Using definition of Laplace transform for each case
(a)
(b)
(c)
(d)
=
+ ( )
92
(e)
= ( 1)
+ (1)
EXAMPLE 6.2
Find the inverse Laplace transform of the following
Solution: Given
It can also be expressed as
( )=5
25 + 1
25 = 5
25 +1
5 5
25
Then the inverse Laplace transform is given by
( )= 5
25 +1
5 5
25 = 5 5 + 1
55
EXAMPLE 6.3
Find the Laplace transform of the following functions.
+
Solution: Using linearity of Laplace transform
EXAMPLE 6.4
Find the Laplace transform of the following function ( ) = .
93
Solution: As we know
Then using first shifting theorem, we get
EXAMPLE 6.5
Find the inverse Laplace transform of the following
Solution: Given
( )= + 1
4 = + 1
( 4)
Using partial fractions
Then the inverse Laplace transform is given by
( )= 1
4 + 5
4( 4) = 1
4+ 5
4
EXAMPLE 6.6
Find the inverse Laplace transform of the following
Solution: As we know
=1
=1
(1 )
94
EXAMPLE 6.7
Solve the IVPs by the Laplace transform.
= , ()= , ()=
Solution: Applying Laplace transform of the given ODE
( 0) ( 0) [ (0)] 6 = 0
( 0) ( 0) + ( 0) 6 = 0
Using given initial conditions,
or,
=11 17
( 6 )= 11 17
( 3 )(+ 2 )= 10
3+ 1
+ 2
Taking using Laplace transform, we get
( )= 10
3+ 1
+ 2 =10 +
EXAMPLE 6.8
Solve the IVPs by the Laplace transform.
+ = , ()= , ( )=
Solution: Applying Laplace transform of the given ODE
( 0) ( 0) 3[ (0)] + 2 =1
+ 4
( 0) ( 0) 3 + 3 (0 )+ 2 =1
+ 4
Using given initial conditions,
5 3 + 3 + 2 =1
+ 4
or,
or,
= + 2
( 3 + 2 )+ 1
( 3 + 2 )(+ 4)
or,
=( + 2)( + 4)+ 1
( 3 + 2 )(+ 4)
95
or,
= + 6 + 9
( 1 )( 2 )(+ 4 )= 16
5( 1)+ 25
6( 2)+ 1
30( + 4)
Taking using Laplace transform, we get
( )= 16
5( 1)+ 25
6( 2)+ 1
30( + 4) = 16
5 + 25
6 + 1
30
EXAMPLE 6.9
Find the Laplace tran sform of f( t ) using the unit step function,
( ) =
< < 1
< < 3
< < 4
> 4
Solution: Given function f (t ) can be expressed in terms of unit step functions as
( )= 1 ( 1) + 2 ( 1) 2 ( 3) + 4 ( 3) 4 ( 4) 2 ( 4)
( )= 1 + ( 1) + 2 ( 3) 6 ( 4)
Then its Laplace transform is given by
{( )} = {1 }+ { ( 1)} + {2 ( 3)} { 6 ( 4)}
( )=1
+
+2
6
EXAMPLE 6.10
Find f (t ) if F (s ) equals
Solution: Given
Using partial fractions
( )=
+
( 1)
Then the inverse Laplace transform is given by
96
( )=
+
( 1 ) = ( 2 ) ( 2 ) ( 2 ) + ( ) ( 2)
EXAMPLE 6.11
Solve the IVPs by the Laplace transform.
+ + = ()+ ( ) , ()= , ()=
Solution: Applying Laplace transform of the given ODE
( 0) ( 0) + 4 [ (0)] +13 = +
( 0) ( 0) + 4 4 (0 )+13 = +
Using given initial conditions,
+ 4 4 + 13 = +
( + 4 +13 )= +4+ +
or,
= + 4
+ 4 +13 + +
+ 4 +13
or,
= + 2
(+ 2 ) + 3 + 2
(+ 2 ) + 3 +
(+ 2 ) + 3 +
(+ 2 ) + 3
Taking using Laplace transform, we get
= + 2
(+ 2 ) + 3 + 2
(+ 2 ) + 3 +
(+ 2 ) + 3 +
(+ 2 ) + 3
= + 2
(+ 2 ) + 3 + 2
3
3
(+ 2 ) + 3 + 1
3
3
(+ 2 ) + 3 + 1
3
3
(+ 2 ) + 3
= 3 + 2
3 3+1
3 ( ) ( ) 3 ( )
+1
3 ( 3 ) ( ) 3 ( 3 )
EXAMPLE 6.12
Find convolution by integration
97
Solution: Using convolution theorem
=
=
EXAMPLE 6.13
Find f (t ) if F (s ) equals
Solution: Given
( )=18
( +36 ) = 3. 6
+ 6 .
+ 6
As we know
and
Then using convolution theorem
98
=3
2 [ 6+ ( 6+12 )]
=3
2 6
3
24 ( 6+12 )
=3
2 6 1
8( 6 6 )
EXAMPLE 6.14
Using the Laplace transform, solve the IVP:
Solution: Applying Laplace transform of the given system of ODEs
Using given initial conditions,
Rearranging the terms
Solving simultaneous equations [a] and [b] , we get
=( 4)
6 +13 = ( 3 1)
( 3 ) + 2 = ( 3)
( 3 ) + 2 1
( 3 ) + 2
99
=5
6 +13 = 5
( 3 ) + 2
Taking the inverse Laplace transform, we get the solution as
= ( 3)
( 3 ) + 2 1
2
2
( 3 ) + 2 = 3 1
2 3
= 5
2
2
( 3 ) + 2 = 5
2 3
EXAMPLE 6.15
Using the Laplace transform, solve the IVP:
Solution: Applying Laplace transform of the given system of ODEs
Using given initial conditions,
Rearranging the terms
[a]
[b]
Solving simultaneous equations [a] and [b] , we get
100
Taking the inverse Laplace transform, we get the solution as
= 1
+ 1 1
+ 1
=
+ 1 + 2
( + 1 )
=
+ 1 + 2
( + 1 ) 1
( + 1 )
= + 2 ( )
101
= + [ + ( 2 + )]
= + + ( 2 + )
2
102
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Applied Mathematics 3 Problems And Solutions Pdf
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