The general solution of the equation, x ′ = ( 2 − 1 3 − 2 ) x + ( e t t ) .

Question

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Answer 1

Question

Chapter 7.9, Problem 1P

To determine

The general solution of the equation, x′=(2−13−2)x+(ett).

Expert Solution & Answer

Answer to Problem 1P

The general solution of x′=(2−13−2)x+(ett) is obtained as, x(t)=c1(13)e−t+c2(11)et+(32t−1432t−34)et+(12)t−(01)(−23et−e−t−13et+23e−t)_.

Explanation of Solution

The given system of equations is,

x′=(2−13−2)x+(ett)

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

Consider the homogeneous part. That is,

x′=(2−13−2)x

This is of the form, x′=Ax, where A=(2−13−2).

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

|A−rI|ξ=0|2−r−13−2−r|(ξ1ξ2)=(00)(2−r)(−2−r)+3=0r2−1=0

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

r2−1=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+1−13−2+1)=(3−13−1)

On solving, (3−13−1)(ξ1ξ2)=(00), the following equation is obtained.

3ξ1−ξ2=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)c2e−t.

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

(2−1−13−2−1)=(1−13−3)

On solving, (1−13−3)(ξ1ξ2)=(00), the following equation is obtained.

ξ1−ξ2=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, T−1=−12(1−1−31).

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+T−1g(t) …(1) where, D=(r100r2)

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

y′=Dy+T−1g(t)(y1′y2′)=(−1001)(y1y2)−12(1−1−31)(ett)

Thus, the following two linear differential equations are obtained.

y1′=−y1−12et+12t …(2)y2′=y2+32et−12t …(3)

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

μ(t)=e∫p(t)dt=e−t

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

y1=c1e−t−14et+12t−12 …(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)(c1e−t−14et+12t−1232tet+12t+12+c2et)=(c1e−t−14et+12t−12+32tet+12t+12+c2et3(c1e−t−14et+12t−12)+32tet+12t+12+c2et)=c1(13)e−t+c2(11)et+(32t−1432t−34)et+(12)t−(01)

Therefore, the general solution of x′=(2−13−2)x+(ett) is obtained as, x(t)=c1(13)e−t+c2(11)et+(32t−1432t−34)et+(12)t−(01)(−23et−e−t−13et+23e−t)_.

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Answer 2

Question

Chapter 7.9, Problem 1P

To determine

The general solution of the equation, x′=(2−13−2)x+(ett).

Expert Solution & Answer

Answer to Problem 1P

The general solution of x′=(2−13−2)x+(ett) is obtained as, x(t)=c1(13)e−t+c2(11)et+(32t−1432t−34)et+(12)t−(01)(−23et−e−t−13et+23e−t)_.

Explanation of Solution

The given system of equations is,

x′=(2−13−2)x+(ett)

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

Consider the homogeneous part. That is,

x′=(2−13−2)x

This is of the form, x′=Ax, where A=(2−13−2).

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

|A−rI|ξ=0|2−r−13−2−r|(ξ1ξ2)=(00)(2−r)(−2−r)+3=0r2−1=0

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

r2−1=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+1−13−2+1)=(3−13−1)

On solving, (3−13−1)(ξ1ξ2)=(00), the following equation is obtained.

3ξ1−ξ2=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)c2e−t.

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

(2−1−13−2−1)=(1−13−3)

On solving, (1−13−3)(ξ1ξ2)=(00), the following equation is obtained.

ξ1−ξ2=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, T−1=−12(1−1−31).

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+T−1g(t) …(1) where, D=(r100r2)

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

y′=Dy+T−1g(t)(y1′y2′)=(−1001)(y1y2)−12(1−1−31)(ett)

Thus, the following two linear differential equations are obtained.

y1′=−y1−12et+12t …(2)y2′=y2+32et−12t …(3)

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

μ(t)=e∫p(t)dt=e−t

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

y1=c1e−t−14et+12t−12 …(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)(c1e−t−14et+12t−1232tet+12t+12+c2et)=(c1e−t−14et+12t−12+32tet+12t+12+c2et3(c1e−t−14et+12t−12)+32tet+12t+12+c2et)=c1(13)e−t+c2(11)et+(32t−1432t−34)et+(12)t−(01)

Therefore, the general solution of x′=(2−13−2)x+(ett) is obtained as, x(t)=c1(13)e−t+c2(11)et+(32t−1432t−34)et+(12)t−(01)(−23et−e−t−13et+23e−t)_.

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Answer 3

Question

Chapter 7.9, Problem 1P

To determine

The general solution of the equation, x′=(2−13−2)x+(ett).

Expert Solution & Answer

Answer to Problem 1P

The general solution of x′=(2−13−2)x+(ett) is obtained as, x(t)=c1(13)e−t+c2(11)et+(32t−1432t−34)et+(12)t−(01)(−23et−e−t−13et+23e−t)_.

Explanation of Solution

The given system of equations is,

x′=(2−13−2)x+(ett)

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

Consider the homogeneous part. That is,

x′=(2−13−2)x

This is of the form, x′=Ax, where A=(2−13−2).

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

|A−rI|ξ=0|2−r−13−2−r|(ξ1ξ2)=(00)(2−r)(−2−r)+3=0r2−1=0

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

r2−1=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+1−13−2+1)=(3−13−1)

On solving, (3−13−1)(ξ1ξ2)=(00), the following equation is obtained.

3ξ1−ξ2=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)c2e−t.

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

(2−1−13−2−1)=(1−13−3)

On solving, (1−13−3)(ξ1ξ2)=(00), the following equation is obtained.

ξ1−ξ2=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, T−1=−12(1−1−31).

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+T−1g(t) …(1) where, D=(r100r2)

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

y′=Dy+T−1g(t)(y1′y2′)=(−1001)(y1y2)−12(1−1−31)(ett)

Thus, the following two linear differential equations are obtained.

y1′=−y1−12et+12t …(2)y2′=y2+32et−12t …(3)

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

μ(t)=e∫p(t)dt=e−t

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

y1=c1e−t−14et+12t−12 …(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)(c1e−t−14et+12t−1232tet+12t+12+c2et)=(c1e−t−14et+12t−12+32tet+12t+12+c2et3(c1e−t−14et+12t−12)+32tet+12t+12+c2et)=c1(13)e−t+c2(11)et+(32t−1432t−34)et+(12)t−(01)

Therefore, the general solution of x′=(2−13−2)x+(ett) is obtained as, x(t)=c1(13)e−t+c2(11)et+(32t−1432t−34)et+(12)t−(01)(−23et−e−t−13et+23e−t)_.

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Answer 4

Question

Chapter 7.9, Problem 1P

To determine

The general solution of the equation, x′=(2−13−2)x+(ett).

Expert Solution & Answer

Answer to Problem 1P

The general solution of x′=(2−13−2)x+(ett) is obtained as, x(t)=c1(13)e−t+c2(11)et+(32t−1432t−34)et+(12)t−(01)(−23et−e−t−13et+23e−t)_.

Explanation of Solution

The given system of equations is,

x′=(2−13−2)x+(ett)

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

Consider the homogeneous part. That is,

x′=(2−13−2)x

This is of the form, x′=Ax, where A=(2−13−2).

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

|A−rI|ξ=0|2−r−13−2−r|(ξ1ξ2)=(00)(2−r)(−2−r)+3=0r2−1=0

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

r2−1=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+1−13−2+1)=(3−13−1)

On solving, (3−13−1)(ξ1ξ2)=(00), the following equation is obtained.

3ξ1−ξ2=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)c2e−t.

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

(2−1−13−2−1)=(1−13−3)

On solving, (1−13−3)(ξ1ξ2)=(00), the following equation is obtained.

ξ1−ξ2=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, T−1=−12(1−1−31).

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+T−1g(t) …(1) where, D=(r100r2)

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

y′=Dy+T−1g(t)(y1′y2′)=(−1001)(y1y2)−12(1−1−31)(ett)

Thus, the following two linear differential equations are obtained.

y1′=−y1−12et+12t …(2)y2′=y2+32et−12t …(3)

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

μ(t)=e∫p(t)dt=e−t

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

y1=c1e−t−14et+12t−12 …(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)(c1e−t−14et+12t−1232tet+12t+12+c2et)=(c1e−t−14et+12t−12+32tet+12t+12+c2et3(c1e−t−14et+12t−12)+32tet+12t+12+c2et)=c1(13)e−t+c2(11)et+(32t−1432t−34)et+(12)t−(01)

Therefore, the general solution of x′=(2−13−2)x+(ett) is obtained as, x(t)=c1(13)e−t+c2(11)et+(32t−1432t−34)et+(12)t−(01)(−23et−e−t−13et+23e−t)_.

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Answer 5

Chapter 7.9, Problem 1P

To determine

The general solution of the equation, x′=(2−13−2)x+(ett).

Expert Solution & Answer

Answer to Problem 1P

The general solution of x′=(2−13−2)x+(ett) is obtained as, x(t)=c1(13)e−t+c2(11)et+(32t−1432t−34)et+(12)t−(01)(−23et−e−t−13et+23e−t)_.

Explanation of Solution

The given system of equations is,

x′=(2−13−2)x+(ett)

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

Consider the homogeneous part. That is,

x′=(2−13−2)x

This is of the form, x′=Ax, where A=(2−13−2).

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

|A−rI|ξ=0|2−r−13−2−r|(ξ1ξ2)=(00)(2−r)(−2−r)+3=0r2−1=0

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

r2−1=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+1−13−2+1)=(3−13−1)

On solving, (3−13−1)(ξ1ξ2)=(00), the following equation is obtained.

3ξ1−ξ2=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)c2e−t.

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

(2−1−13−2−1)=(1−13−3)

On solving, (1−13−3)(ξ1ξ2)=(00), the following equation is obtained.

ξ1−ξ2=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, T−1=−12(1−1−31).

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+T−1g(t) …(1) where, D=(r100r2)

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

y′=Dy+T−1g(t)(y1′y2′)=(−1001)(y1y2)−12(1−1−31)(ett)

Thus, the following two linear differential equations are obtained.

y1′=−y1−12et+12t …(2)y2′=y2+32et−12t …(3)

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

μ(t)=e∫p(t)dt=e−t

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

y1=c1e−t−14et+12t−12 …(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)(c1e−t−14et+12t−1232tet+12t+12+c2et)=(c1e−t−14et+12t−12+32tet+12t+12+c2et3(c1e−t−14et+12t−12)+32tet+12t+12+c2et)=c1(13)e−t+c2(11)et+(32t−1432t−34)et+(12)t−(01)

Therefore, the general solution of x′=(2−13−2)x+(ett) is obtained as, x(t)=c1(13)e−t+c2(11)et+(32t−1432t−34)et+(12)t−(01)(−23et−e−t−13et+23e−t)_.

Answer 6

Chapter 7.9, Problem 1P

To determine

The general solution of the equation, x′=(2−13−2)x+(ett).

Expert Solution & Answer

Answer to Problem 1P

The general solution of x′=(2−13−2)x+(ett) is obtained as, x(t)=c1(13)e−t+c2(11)et+(32t−1432t−34)et+(12)t−(01)(−23et−e−t−13et+23e−t)_.

Explanation of Solution

The given system of equations is,

x′=(2−13−2)x+(ett)

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

Consider the homogeneous part. That is,

x′=(2−13−2)x

This is of the form, x′=Ax, where A=(2−13−2).

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

|A−rI|ξ=0|2−r−13−2−r|(ξ1ξ2)=(00)(2−r)(−2−r)+3=0r2−1=0

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

r2−1=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+1−13−2+1)=(3−13−1)

On solving, (3−13−1)(ξ1ξ2)=(00), the following equation is obtained.

3ξ1−ξ2=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)c2e−t.

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

(2−1−13−2−1)=(1−13−3)

On solving, (1−13−3)(ξ1ξ2)=(00), the following equation is obtained.

ξ1−ξ2=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, T−1=−12(1−1−31).

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+T−1g(t) …(1) where, D=(r100r2)

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

y′=Dy+T−1g(t)(y1′y2′)=(−1001)(y1y2)−12(1−1−31)(ett)

Thus, the following two linear differential equations are obtained.

y1′=−y1−12et+12t …(2)y2′=y2+32et−12t …(3)

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

μ(t)=e∫p(t)dt=e−t

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

y1=c1e−t−14et+12t−12 …(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)(c1e−t−14et+12t−1232tet+12t+12+c2et)=(c1e−t−14et+12t−12+32tet+12t+12+c2et3(c1e−t−14et+12t−12)+32tet+12t+12+c2et)=c1(13)e−t+c2(11)et+(32t−1432t−34)et+(12)t−(01)

Therefore, the general solution of x′=(2−13−2)x+(ett) is obtained as, x(t)=c1(13)e−t+c2(11)et+(32t−1432t−34)et+(12)t−(01)(−23et−e−t−13et+23e−t)_.

Answer 7

Chapter 7.9, Problem 1P

To determine

The general solution of the equation, x′=(2−13−2)x+(ett).

Answer 8

Expert Solution & Answer

Answer to Problem 1P

The general solution of x′=(2−13−2)x+(ett) is obtained as, x(t)=c1(13)e−t+c2(11)et+(32t−1432t−34)et+(12)t−(01)(−23et−e−t−13et+23e−t)_.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

The given system of equations is,

x′=(2−13−2)x+(ett)

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

Consider the homogeneous part. That is,

x′=(2−13−2)x

This is of the form, x′=Ax, where A=(2−13−2).

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

|A−rI|ξ=0|2−r−13−2−r|(ξ1ξ2)=(00)(2−r)(−2−r)+3=0r2−1=0

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

r2−1=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+1−13−2+1)=(3−13−1)

On solving, (3−13−1)(ξ1ξ2)=(00), the following equation is obtained.

3ξ1−ξ2=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)c2e−t.

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

(2−1−13−2−1)=(1−13−3)

On solving, (1−13−3)(ξ1ξ2)=(00), the following equation is obtained.

ξ1−ξ2=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, T−1=−12(1−1−31).

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+T−1g(t) …(1) where, D=(r100r2)

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

y′=Dy+T−1g(t)(y1′y2′)=(−1001)(y1y2)−12(1−1−31)(ett)

Thus, the following two linear differential equations are obtained.

y1′=−y1−12et+12t …(2)y2′=y2+32et−12t …(3)

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

μ(t)=e∫p(t)dt=e−t

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

y1=c1e−t−14et+12t−12 …(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)(c1e−t−14et+12t−1232tet+12t+12+c2et)=(c1e−t−14et+12t−12+32tet+12t+12+c2et3(c1e−t−14et+12t−12)+32tet+12t+12+c2et)=c1(13)e−t+c2(11)et+(32t−1432t−34)et+(12)t−(01)