# Solving Linear Differential Equation

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 $y_2(t)$?

-
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