I'm trying to find the particular solution $y_p$ for:

$$y''+y=\sin x$$

I set $y_p(x)=A\sin x + B\cos x$ and differentiate 2 times: $$y_p'(x)=A\cos x - B\sin x$$ $$y_p''(x)=-A\sin x - B\cos x$$ I Insert into $$y_p''+y_p=\sin x$$

$$-A\sin x - B\cos x+ A\sin x + B\cos x=\sin x$$

And this cant be right, because the LHS. is $=0$

Where did I go wrong?

  • 3
    $\begingroup$ Thats because your $y_p$ is solution to the homogenous $y''+y=0$. $\endgroup$ – Yadati Kiran Nov 25 '18 at 16:30
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    $\begingroup$ The method of undetermined coefficients demands that you add a factor $x$ in the resonance case (with multiplicity one), $y_p(x)=Ax\sin x+Bx\cos x.$ $\endgroup$ – Lutz Lehmann Nov 25 '18 at 16:36
  • $\begingroup$ @Curl: In such cases you may take $y_p=Ax\sin x$ alone (if you are adventurous). $\endgroup$ – Yadati Kiran Nov 25 '18 at 16:48
  • $\begingroup$ @LutzL Nice, this works. $\endgroup$ – Curl Nov 25 '18 at 16:48
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    $\begingroup$ @Isham: That's why I said "if you are adventurous" (i.e. if you realise you can't if you have terms in $\cos x$). $\endgroup$ – Yadati Kiran Nov 25 '18 at 17:18

When the RHS has the form $$P_n(x)\sin(\omega x)$$

and $(i\omega)$ is a root of the caracteristic equation, the particular solution of the equation $$y''+y=P_n(x)\sin(\omega x)$$

will be as $$y_p=\color{red}{xQ_n(x)}\Bigl(A\cos(\omega x)+B\sin(\omega x)\Bigr)$$

$P_n$ and $Q_n$ are polynomial of degree $=n$.

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