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

Nov 2009
717
133
Wahiawa, Hawaii
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
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
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.
 
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romsek

MHF Helper
Nov 2013
6,666
3,004
California
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.
 
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