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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 $y_2(t)$?

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    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](http://math.stackexchange.com/questions/33851/how-to-calculate-the-matrix-exponential-explicitly-for-a-matrix-which-isnt-diago/34139#34139) for instance.2011-09-20

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