Elementary Differential Equations
Elementary Differential Equations
10th Edition
ISBN: 9780470458327
Author: William E. Boyce, Richard C. DiPrima
Publisher: Wiley, John & Sons, Incorporated
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Chapter 7.9, Problem 1P
To determine

The general solution of the equation, x=(2132)x+(ett).

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Answer to Problem 1P

The general solution of x=(2132)x+(ett) is obtained as, x(t)=c1(13)et+c2(11)et+(32t1432t34)et+(12)t(01)(23etet13et+23et)_.

Explanation of Solution

The given system of equations is,

x=(2132)x+(ett)

It is noted that, the given system is a non-homogeneous system.

Consider the homogeneous part. That is,

x=(2132)x

This is of the form, x=Ax, where A=(2132).

The eigenvalues r and the eigenvectors ξ satisfy the equation, |ArI|ξ=0. That is,

|ArI|ξ=0|2r132r|(ξ1ξ2)=(00)(2r)(2r)+3=0r21=0

The roots of the above characteristic equation is obtained as follows.

r21=0r2=1

Thus, the solutions are, r1=1,r2=1. That is, the solutions are real and different.

Set r1=1 in the coefficient matrix. That is,

(2+1132+1)=(3131)

On solving, (3131)(ξ1ξ2)=(00), the following equation is obtained.

3ξ1ξ2=0

Let ξ1=1, then the above equation becomes, ξ2=3.

Then, the eigen vector formed is, ξ=(13).

Thus, the first solution of the homogeneous solution is, ξ(1)=(13)c2et.

Set r2=1 in the coefficient matrix. That is,

(211321)=(1133)

On solving, (1133)(ξ1ξ2)=(00), the following equation is obtained.

ξ1ξ2=0ξ1=ξ2

Let ξ1=1, then from the above equation, ξ2=1.

Then, the eigen vector formed is, ξ=(11).

Thus, the second solution of the homogeneous solution is, ξ(2)=c1(11)et.

Therefore, the matrix T of eigen vectors is,

T=(ξ(1),ξ(2))=(1131)

Then, the inverse of T is, T1=12(1131).

It is noted from the given equation that, g(t)=(ett).

Now consider the equation, x=Ty. Then, the following equation is obtained.

y=Dy+T1g(t)  (1) where, D=(r100r2)

Substitute the eigen values and inverse of T in the equation (1) as follows.

y=Dy+T1g(t)(y1y2)=(1001)(y1y2)12(1131)(ett)

Thus, the following two linear differential equations are obtained.

y1=y112et+12t    (2)y2=y2+32et12t      (3)

Consider the equation (2). The integral factor is,

μ(t)=ep(t)dt=et

Multiply the obtained integral factor throughout the equation (2) and y1 is obtained as,

y1=c1et14et+12t12      (4)

Similarly, multiply the obtained integral factor et throughout the equation (3) and y2 is obtained as,

y2=32tet+12t+12+c2et     (5)

The solution is given by the equation, x=Ty. That is, x(t)=T(y1y2)       (6).

Substitute the obtained solutions in the equation (6) as follows.

x(t)=(1131)(y1y2)=(1131)(c1et14et+12t1232tet+12t+12+c2et)=(c1et14et+12t12+32tet+12t+12+c2et3(c1et14et+12t12)+32tet+12t+12+c2et)=c1(13)et+c2(11)et+(32t1432t34)et+(12)t(01)

Therefore, the general solution of x=(2132)x+(ett) is obtained as, x(t)=c1(13)et+c2(11)et+(32t1432t34)et+(12)t(01)(23etet13et+23et)_.

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Chapter 7 Solutions

Elementary Differential Equations

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