Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have the following : $$ Y'(t)=JY(t)+K(t) $$ where $Y, J, K$ are matrices. The text book I have tells me the following for $J=J_3(\lambda)$: $$\begin{align} y'_1(t)&=\lambda y_1(t)+K_1(t)\\ y'_2(t)&=\lambda y_2(t)+(y_1(t)+K_2(t))\\ y'_3(t)&=\lambda y_3(t)+(y_2(t)+K_3(t)) \end{align}$$

If I have $$K(t)= \begin{pmatrix} 3\\ e^t-6\\ e^{-3t}-e^t+3 \end{pmatrix}$$

then am I correct in setting up the first equation as follows: If $\lambda=-1$, $y'_1(t)=-y_1(t)+3$, then I could use the integrating factor of $e^t$ and solve to get: $y_1$(t)=3+c$e^{-t}$

Is this correct so far?

Can someone please step me through getting $y_2(t)$?

share|cite|improve this question
Are you sure that your textbook wants you to solve the equations one by one but not with a Laplace transform? – user13838 Sep 20 '11 at 11:21
and also J is for a possible Jordan block? – user13838 Sep 20 '11 at 11:51
The solution to $Y'=AY+K(t)$, $Y(t_0)=Y_0$, where $A$ is a constant square matrix and $K$ is continuous, is $$e^{(t-t_0)A}\ Y_0+\int_{t_0}^te^{(t-x)A}\ K(x)\ dx.$$ Moreover, there is a closed formula for $e^{tA}$: see this answer for instance. – Pierre-Yves Gaillard Sep 20 '11 at 14:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.