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

#### karush

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

#### karush

##### Well-known member
wow, that was a really a great help...

not sure why the latex didn't render completely