# How to solve ODE with cosine involved

I sought for this really long in the internet but didn't happen to find a general solution. So here is the equation:

$$y' = y + x\,\cos(2\,x)$$

I know that for the particular solution $$y_p$$ following assumption is made:

$$y_p = (a_0+a_1\,x)\,\sin(2\,x)+(b_0+b_1\,x)\,\cos(2\,x)$$

so that

\begin{align} {y_p}' &= a_1\,\sin(2\,x)+(a_0+a_1\,x)\,2\,\cos(2\,x)+ b_1\,\cos(2\,x)-(b_0+b_1\,x)\,2\,\sin(2\,x) \\ &=y + x\,\cos(2\,x) = {y_p}' \end{align}

But I don't know at all how to determine $$a_1, a_0, b_1, b_0$$

I could also just solve $$y = C\,e^{x} +e^{x}\,\int{\frac{\cos(2\,x)}{e^x}}\mathrm{d}x$$ thanks to another answer here, but how to do it generally?

• I mean the computation that I quitted halfway. So solving it explicitly for the constants in order to understand It generally would be great.
– Leon
Jan 2, 2021 at 17:25
• Plug in $y_p$ for $y$ on the r.h.s. and group like terms and set them equal.
– IanJ
Jan 2, 2021 at 17:44
• The equation has to be true for all $x$, so you have to set the coefficients of the functions equal to each other on both sides, i.e. the coefficent of $x\cos x$ on the left equals the coefficient of $x\cos x$ on the right and so on. Jan 2, 2021 at 17:45

Grouping by $$sin(2x)$$ gives the equation:

$$a_1 - 2b_0 = a_0$$

Grouping by $$x sin(2x)$$:

$$-2b_1 = a_1$$

Grouping by $$cos(2x)$$:

$$2a_0 + b_1 = b_0$$

Grouping by $$x cos(2x)$$: $$2a_1 = b_1 + 1$$

Solve and you get $$b_1 = -1/5$$, $$a_1 = 2/5$$, $$a_0 = 4/25$$, $$b_0 = 3/25$$.

• Thank you so much!
– Leon
Jan 2, 2021 at 20:08