I have found the fundamental solution set of the LHS to be $J_1=cos(x)$, $J_2=sin(x)$, but then I find the solution set of the RHS to be $J_3=xcos(x)$, $J_4=xsin(x)$. If I then apply the operator $A=D^2+1$ to this solution set to find a particular solution I find $A(c_1*J_3+c_2*J_4)=(-2c_1+c_2x)*sin(x)+(-c_1x+2c_2)cos(x)=sin(x)$, which makes it impossible to find a constant value for $c_1$ and $c_2$. I am probably doing something wrong here, but I can't seem to find my error.

Edit: I did find my error, I forgot adding f at the end and only focussed on the f'' part. Thanks for helping me!

  • $\begingroup$ $AJ_3=-2\sin x$, $AJ_4=2\cos x$. Yes, you're doing something wrong. Write your result of $J_3'$ and $J_3''$ please. $\endgroup$ Oct 25, 2021 at 14:46
  • $\begingroup$ Your computation of the second derivative is incomplete. $(x\sin x)''=-x\sin x+2\cos x$ etc. $\endgroup$ Oct 25, 2021 at 14:46
  • $\begingroup$ The method of undetermined coefficients is okay, but I suggest (if you haven't tried it already) that you look into the method of annihilators for problems like this. It might be a little less terse, but it's conceptually way clearer. $\endgroup$ Oct 25, 2021 at 15:27

1 Answer 1


$$f''+f=\sin(x).$$ The guess should simply be: $$f_p=x(C_1 \sin x +C_2 \cos x)$$ $$f_p=C_1 x \sin x +C_2 x\cos x$$ But you don't really need the $C_1x\sin x$ part. Since $C_1$ should be equal to zero.

So just try: $f_p=C_2 x\cos x$.

$$\implies f'_p=C_2 \cos x -C_2 x \sin x$$ $$f''_p=-2C_2 \sin x -C_2x \cos x$$ $$f''_p+f_p=-2C_2 \sin x =\sin x$$ $$\implies C_2=-\dfrac 12 $$ Therefore: $$f(x)=A \sin x +B \cos x -\dfrac 12 x \cos x.$$

  • $\begingroup$ I'm sorry, I think I was taught a different notation than is usual, but what I am trying to do is then find the A in your equation. When I try to do so, I find an equation with A and x, where I would normally expect to find one just with A. $\endgroup$ Oct 25, 2021 at 14:48
  • $\begingroup$ The guess normally should be $f_p=C_1 x \sin x +C_2 x\cos x$ but then $C_1=0$ b @davidhintzen $\endgroup$ Oct 25, 2021 at 14:49

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