I thought I grasped it thanks to an answer to a similar question I posed, but it seems a little trickier:

Given $y' = y\,\sin(x)+\sin(2\,x)$, trying to solve with the same approach:

$y_p = a_0\,\sin(2\,x)+b_0\,\cos(2\,x)$

so that:

$\begin{align} {y_p}' &= 2\,a_0\cos(2\,x) -2\,b_0\,\sin(2\,x)\\ &= y\,\sin(x)+\sin(2\,x) \\ &= a_0\,\sin(2\,x)\,\sin(x)+b_0\,\cos(2\,x)\,\sin(x) + \sin(2\,x) \end{align}$

Unlike the other inquiry no comparison of coefficients seems possible due to the multiplication.

Is there something to be done still?


1 Answer 1


The method of undetermined coefficients works for linear ODE with constant coefficients. The given ODE is a linear ODE with variable coefficients.

After multiplying both sides by the integral factor $e^{\cos(x)}$, the given linear ODE is equivalent to $$D(e^{\cos(x)}y(x))=e^{\cos(x)}\sin(2x).$$ Now integrate the RHS: $$\begin{align} \int e^{\cos(x)}\sin(2x)\,dx&=2\int e^{\cos(x)}\cos(x)\sin(x)\,dx\\ &=-2\int e^{t}t\,dt=-2e^t(t-1)+c\\ &=-2e^{\cos(x)}(\cos(x)-1)+c. \end{align}$$ Hence the general solution is $$y(x)=ce^{-\cos(x)}+\underbrace{2(1-\cos(x))}_{\text{particular solution}}.$$ Note that the particular solution $y_p(x)=2(1-\cos(x))$ is not of the form: $a_0\sin(2x)+b_0\cos(2x)$. Indeed $y_p(0)=0=b_0$ and $y_p(\pi)=4=b_0$ which is a contradiction.

  • $\begingroup$ wow, that's so shrewd, thank you! $\endgroup$
    – Leon
    Commented Jan 3, 2021 at 12:12

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