# differential equations for partial solution

Solve the system $$x'= \left(\begin{matrix}1&1&0\\1&1&0\\0&0&3\end{matrix}\right)x +\left(\begin{matrix}e^t\\e^{2t}\\3e^{3t}\end{matrix}\right)$$ I got the eigenvalues $\lambda_1=0; \lambda_2=2; \lambda_3=3$ and the solution for homogeneous part is $$x=C_1\left(\begin{matrix}-1\\1\\0\end{matrix}\right)+C_2e^{2t}\left(\begin{matrix}1\\1\\0\end{matrix}\right)+C_3e^{3t}\left(\begin{matrix}1\\2\\1\end{matrix}\right)$$ I am stuck here. I don't know how to find the partial solution, please help

• @user73192: Did they provide initial conditions at say $t = 0), or even say$x(0) = x_0$? – Amzoti Aug 19 '13 at 0:08 ## 1 Answer I will provide a guiding hint. Give it a go and respond back if lost. Given: $$x'= Ax + F(t) = \left[\begin{matrix}1&1&0\\1&1&0\\0&0&3\end{matrix}\right]x +\left[\begin{matrix}e^t\\e^{2t}\\3e^{3t}\end{matrix}\right]$$ So,$ A = \left[\begin{matrix}1&1&0\\1&1&0\\0&0&3\end{matrix}\right]$and$F(t) = \left[\begin{matrix}e^t\\e^{2t}\\3e^{3t}\end{matrix}\right]$• We want to find the matrix exponential$e^{At}$• We then solve for$X(t) = e^{At}x(\tau) + \int_\tau^t e^{A(t-s)}F(s)ds$, where$\tau$is the initial condition time, but you do not have that, so we need to leave it in a general form. In other words$x(\tau) = C$, some set of initial values for some given$t$. So, the matrix exponential is given by (this is different than what you got): $$e^{At} = \left[\begin{matrix} \dfrac{1}{2} (1 + e^{2 t}) & \dfrac{1}{2} (-1 + e^{2 t}) & 0 \\ \dfrac{1}{2} (-1 + e^{2 t}) & \dfrac{1}{2} (1 + e^{2 t}) & 0 \\ 0 & 0 & e^{3t} \end{matrix}\right]$$ • Next, find$e^{A(t-s)}F(s)ds$, so we have (just multiply these two matrices): •$e^{A(t-s)}F(s) = \left[\begin{matrix} \dfrac{1}{2} (1 + e^{2 (t-s)}) & \dfrac{1}{2} (-1 + e^{2(t-s)}) & 0 \\ \dfrac{1}{2} (-1 + e^{2 (t-s)}) & \dfrac{1}{2} (1 + e^{2 (t-s)}) & 0 \\ 0 & 0 & e^{3(t-s)} \end{matrix}\right] \cdot \left[\begin{matrix}e^s\\e^{2s}\\3e^{3s}\end{matrix}\right]$• Lastly, find$X(t) = e^{At}x(\tau) + \int_\tau^t e^{A(t-s)}F(s)ds\$

• Nice work again, as usual! +1 – amWhy Aug 18 '13 at 23:58