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[SOLVED] 29.1 Find a general solution to the system of differential equations

karush

Well-known member
Jan 31, 2012
2,886
Find a general solution to the system of differential equations
\begin{align*}\displaystyle
y'_1&=2y_1+3y_2+5x\\
y'_2&=y_1+4y_2+10
\end{align*}
rewrite as

$$Y'=\left[\begin{array}{c}2 & 3 \\ 1 & 4 \end{array}\right]Y
+\left[\begin{array}{c}5x\\ 10\end{array}\right]$$

ok not sure what to do with this
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,181
Find a general solution to the system of differential equations
\begin{align*}\displaystyle
y'_1&=2y_1+3y_2+5x\\
y'_2&=y_1+4y_2+10
\end{align*}
rewrite as

$$Y'=\left[\begin{array}{c}2 & 3 \\ 1 & 4 \end{array}\right]Y
+\left[\begin{array}{c}5x\\ 10\end{array}\right]$$

ok not sure what to do with this
Just like as a single differential equation, solve this first. (I don't know the correct terminology here, but I would call this the homogeneous solution.)
\(\displaystyle \left ( \begin{matrix} y_1 \\ y_2 \end{matrix} \right ) ^{\prime} = \left [ \begin{matrix} 2 & 3 \\ 1 & 4 \end{matrix} \right ] ~ \left ( \begin{matrix} y_ 1 \\ y_2 \end{matrix} \right )\)

Big hint: What are the eigenvalues of the matrix?

-Dan
 

karush

Well-known member
Jan 31, 2012
2,886
Just like as a single differential equation, solve this first. (I don't know the correct terminology here, but I would call this the homogeneous solution.)
\(\displaystyle \left ( \begin{matrix} y_1 \\ y_2 \end{matrix} \right ) ^{\prime} = \left [ \begin{matrix} 2 & 3 \\ 1 & 4 \end{matrix} \right ] ~ \left ( \begin{matrix} y_ 1 \\ y_2 \end{matrix} \right )\)

Big hint: What are the eigenvalues of the matrix?

-Dan

$\left[\begin{array}{c}2 & 3 \\ 1 & 4 \end{array}\right]$
$\left| \begin{array}{cc} - \lambda + 2 & 3 \\1 & - \lambda + 4 \end{array} \right|
=\left(- \lambda + 2\right) \left(- \lambda + 4\right) - 3
=\lambda^2-6\lambda+8-3
=(\lambda-5)(\lambda-1)=0$
so the zeros are
$\lambda = 5,1$
well so far
ok I quess we don't need the eiganvectors?
 
Last edited:

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Yes, you do need the eigenvectors. An eigenvector corresponding to eigenvalue 1 satisifies [itex]\begin{bmatrix}2 & 3 \\ 1 & 4\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}x \\ y \end{bmatrix}[itex] which is equivalent to the two equations 2x+3y= x and x+ 4y= y. Those are both equivalent to x+ 3y= 0 or x= -3y. All eigenvectors corresponding to eigenvalue 1 are of the form [itex]\begin{bmatrix}-3y \\ y \end{bmatrix}[/itex]. Taking y= 1 an eigenvector is [itex]\begin{bmatrix}-3 \\ 1 \end{bmatrix}[/itex].
An eigenvector corresponding to eigenvalue 5 satisifies [itex]\begin{bmatrix}2 & 3 \\ 1 & 4\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}5x \\ 5y \end{bmatrix}[itex] which is equivalent to the two equations 2x+3y= 5x and x+ 4y= 5y. Those are both equivalent to x= y. All eigenvectors corresponding to eigenvalue 5 are of the form [itex]\begin{bmatrix} y\\ y \end{bmatrix}[/itex]. Taking y= 1 an eigenvector is [itex]\begin{bmatrix}1 \\ 1 \end{bmatrix}[/itex].
And the reason we need the eigenvectors is this: Let M be the matrix having those eigenvectors as columns: [itex]M= \begin{bmatrix}-3 & 1 \\ 1 & 1 \end{bmatrix}[/itex] and [itex]M^{-1}= \begin{bmatrix}-\frac{1}{4} & \frac{1}{4} \\ \\frac{1}{4} & \frac{3}{4}\end{bmatrix}[/itex].
Then [itex]\begin{bmatrix}-\frac{1}{4} & \frac{1}{4} \\ \frac{1}{4} & \frac{3}{4}\end{bmatrix}\begin{bmatrix}2 & 3 \\ 1 & 4\end{bmatrix}\begin{bmatrix}-\frac{1}{4} & \frac{1}{4} \\ \\frac{1}{4} & \frac{3}{4}\end{bmatrix}= \begin{bmatrix}1 & 0 \\ 0 & 5\end{bmatrix}[/itex], the diagonal matrix having the eigenvalues of the original matrix on the diagonal. (Any n by n matrix having n independent eigenvectors can be "diagonalized" that way.)
To use that, think of the general differential equation Y'= AY+ B, such that [itex]M^{-1}AM= D[/itex] with D the diagonal matrix having the eigenvalues of A on the diagonal and M the matrix with the eigenvectors of A as columns. Let [itex]U= M^{-1}Y[/itex] so that [itex]Y= MU[/itex]. Since M is a constant matrix we can write the equation as (MU)'= MU'= AMU. Multiply by [itex]M^{-1}[/itex]: [itex]U'= M^{-1}AMU= DU[/itex]. Since D is a diagonal matrix that separates into separate equations: [itex]M^{-1}Y'= M^{-1}AMY+ M^{-1}B[/itex[ or [itex]U'= DY+ M^{-1}B[/itex].
In our case, the matrix equation becomes [itex]U'= \begin{bmatrix}u' \\ v' \end{bmatix}= \begin{bmatrix}1 & 0 \\ 0 & 5 \end{bmatrix}\begin{bmatrix}u \\ v \end{bmatrix}= \begin{bmatrix}u \\ 5v\end{bmatrix}+ \begin{bmatrix}-\frac{5x}{4}+ \frac{5}{2} \\ \frac{25x}{4}+ \frac{75}{4}\end{bmatrix}[/itex].
That separates into the equations [itex]u'= u- \frac{5x}{4}+ \frac{5}{2}[/itex] which has general solution [itex]u(x)= C_1e^{x}- \frac{5x}{4}- \frac{15}{4}[/itex] and [itex]v'= 5v+ \frac{25x}{4}+ \frac{75}{4}[/itex] which has general solution [itex]v(x)= C_2e^{5x}-\frac{5x}{4}+ 4[/itex]. So [itex]U= \begin{bmatrix}C_1e^x- \frac{5x}{4}-\frac{15}{4} \\ C_2e^{5x}- \frac{5x}{4}+ 4\end{bmatrix}[/itex]. Since [itex]U= M^{-1}Y[/itex], [itex]Y= MU= \begin{bmatrix}-3C_1e^x+ C_2e^{5x}+ \frac{15x}{2}+ 4 \\ C_1e^x+ C_2e^{5x}- \frac{5x}{2}+ \frac{1}{4}[/itex]
(modulo any arithmetic errors!)
 

karush

Well-known member
Jan 31, 2012
2,886
wow, that was a really a great help...

not sure why the latex didn't render completely