Thread: s3.16.1 Find the general solution to the system of differential equations

1. s3.16.1 Find the general solution to the system of differential equations

Find the general solution to the system of differential equations

$$\begin{cases} y'_1&=2y_1+y_2-y_3 \\y'_2&=3y_2+y_3\\y'_3&=3y_3 \end{cases}$$

ok I assume the next step is

$$\begin{bmatrix} y'_1 \\y'_2 \\y'_3 \end{bmatrix}=\begin{bmatrix} 2 & 1 & -1 \\0 & 3 & 1 \\ 0 & 0 & 3 \end{bmatrix} \begin{bmatrix} y_1 \\y_2 \\y_3 \end{bmatrix}$$

I had this posted on another forum but think I was not starting out right so got no reply's

2. Re: s3.16.1 Find the general solution to the system of differential equations

I would not use matrices at all. The third equation, $\displaystyle y_3'= 3y_3$, involves only $\displaystyle y_3$ and can be integrated directly: $\displaystyle y_3(t)= c_3e^{3t}$.

The second equation is $\displaystyle y_2'= 3y_2+ y_3= 3y_2+ c_3e^{3t}$. That is a "non-homogeneous equation with constant equations". We can first solve the "associated homogeneous equation", $\displaystyle y_2'= 3y_2$. The general solution to that is $\displaystyle y= c_2e^{3t}$ just as before. We also seek a single solution to the entire equation. Normally, since the "non-homogeneous" part is $\displaystyle c_3e^{3t}$ we would try a multiple of that but that is already a solution to the homogeneous equation so we try $\displaystyle y= Ac_3te^{3t}$. Then $\displaystyle y'= Ac_3e^{3t}+ 3Ac_3te^{3t}$ so the equation becomes $\displaystyle Ac_3e^{3t}+ 3Ac_3te^{3t}= Ac_3te^{3t}+ c_3e^{3t}$. That reduces to $\displaystyle Ac_3e^{3t}= c_3e^{3t}$ so A= 1.
$\displaystyle y_2= c_2e^{3t}+ c_3te^{3t}$.

Now, the first equation is $\displaystyle y_1'= 2y_1+ y_2- y_3= 2y_1+ c_2e^{3t}+ 2c_3te^{3t}- c_3e^{3t}$. The associated homogeneous equation is $\displaystyle y_1'= 2y_1$ which has general solution $\displaystyle y= c_1e^{2t}$. To find a solution to the entire equation try $\displaystyle y= (Ax+ B)e^{3t}$ and find values for A and B that make that true.

3. Re: s3.16.1 Find the general solution to the system of differential equations Originally Posted by HallsofIvy I would not use matrices at all.
But I suspect the point of the exercise is to know how to deal with repeated eigenvalues and rank deficient eigenvectors.