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

Question

Want to see more full solutions like this?

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, $x(t)=c1(13)e−t+c2(11)et+(32t−1432t−34)et+(12)t−(01)(−23et−e−t−13et+23e−t)_$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

image below

Pls help ASAP both

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, $x(t)=c1(13)e−t+c2(11)et+(32t−1432t−34)et+(12)t−(01)(−23et−e−t−13et+23e−t)_$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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, $x(t)=c1(13)e−t+c2(11)et+(32t−1432t−34)et+(12)t−(01)(−23et−e−t−13et+23e−t)_$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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, $x(t)=c1(13)e−t+c2(11)et+(32t−1432t−34)et+(12)t−(01)(−23et−e−t−13et+23e−t)_$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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.

$x(t)=(1131)(y1y2)=(1131)(c1e−t−14et+12t−1232tet+12t+12+c2et)=(c1e−t−14et+12t−12+32tet+12t+12+c2et3(c1e−t−14et+12t−12)+32tet+12t+12+c2et)=c1(13)e−t+c2(11)et+(32t−1432t−34)et+(12)t−(01)$

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 12

Ch. 7.1 - In each of Problems 1 through 4, transform the...Ch. 7.1 - Prob. 2P Ch. 7.1 - Prob. 3P Ch. 7.1 - In each of Problems 1 through 4, transform the...Ch. 7.1 - Prob. 5P Ch. 7.1 - In each of Problems 5 and 6, transform the given...Ch. 7.1 - Systems of first order equations can sometimes be...Ch. 7.1 - Prob. 8P Ch. 7.1 - Prob. 9P Ch. 7.1 - In each of Problems 8 through 12, proceed as in...

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

Videos

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

Answer to Problem 1P

Explanation of Solution

Want to see more full solutions like this?

Chapter 7 Solutions