I use undetermined coefficient method to find that the solution is





However,I see the solution in my book and I don't quite understanding its method

It separates the R.H.S into 2 parts

Let $y_p=Ae^x\cos2x+Be^x\sin2x$ first

$$\begin{pmatrix}1 &4 \\ -4 &1 \end{pmatrix} \begin{pmatrix}A\\ B\end{pmatrix} = \begin{pmatrix}1\\ 0\end{pmatrix}$$



Let $y_{p2}=Ce^{3x}$



How can I to do it in this way?


You can do that with every linear ODE. If $L(y)=0$ is the homogeneous equation, with $L$ as the differential operator, here $L=\frac{d^2}{dx^2}+4$, and you have particular solutions for each of the terms in the sum on the right side, $L(y_{p1})=f_1$, $L(y_{p2})=f_2$, …, $L(y_{pk})=f_k$, then by linearity you can assemble a particular solution for the full sum on the right side as $$ L(y_{p1}+y_{p2}+…+y_{pk})=f_1+f_2+…+f_k. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.