Given that the BVP is a second-order inhomogeneous ODE, we find the characteristic equation to be in the form $$r^2+a^2=0\implies r=\pm ai.$$

Thus, the homogeneous equation is $$u_h=C_1\cos(ax)+C_2\sin(ax).$$

The particular equation is $$u_p=C_3x\sin(\pi x)+C_4x\cos(\pi x),$$ $$u_p'=-C_3\pi x\cos(\pi x)+C_3\sin(\pi x)-C_4\pi x\sin(\pi x)+C_4\cos(\pi x),$$ $$u_p''=-C_3\pi^2 x\sin(\pi x)+2C_3\pi\cos(\pi x)-2C_4\pi\sin(\pi x)-C_4\pi^2 x\cos(\pi x).$$

Substituting for the given ODE, we have $$-C_3\pi^2 x\sin(\pi x)+2C_3\pi\cos(\pi x)-2C_4\sin(\pi x)-C_4\pi^2 x\cos(\pi x)+\pi^2(C_3x\sin(\pi x)+C_4x\cos(\pi x))=\sin(\pi x),$$

and we get $C_3=0$ and $C_4=-\frac{1}{2\pi}$. Then the general solution is now, $$u=u_h+u_p=C_1\cos(ax)+C_2\sin(ax)-\frac{1}{2\pi}x\cos(\pi x)$$

Using the initial conditions, we get $C_1=1$ and $C_2=\frac{-2-\frac{1}{2\pi}-\cos(a)}{\sin(a)}$, but $a=\pm\pi$ and so $\sin(\pm\pi)=0$, which indicates $u$ is undefined when $a=\pm\pi$. But is it truly undefined? I am not sure I solved this correctly.

  • $\begingroup$ What do u mean by BVP? $\endgroup$ – Shobhit Feb 23 '17 at 3:46
  • $\begingroup$ @Shobhit, boundary value problem. $\endgroup$ – ozarka Feb 23 '17 at 3:49

It is impossible to show you exactly where you made a mistake because in the wording of your question, you didn't write down the ODE, nor the initial conditions.

Supposing that the ODE is : $$y''+a^2y=\sin(\pi x)$$ The general solution is : $$y(x)=C_1\cos(ax)+C_2\sin(ax)+\frac{1}{\pi^2-a^2}\sin(\pi x)\qquad\text{if } a\neq \pm \pi$$ or $$y(x)=C_1\cos(\pi x)+C_2\sin(\pi x)-\frac{1}{2\pi}x\cos(\pi x)\qquad\text{if } a= \pm \pi$$ If the first condition is $y(0)=0$ , then $C_1=0 $

Without knowing exactly what is the second condition, one cannot go further.


Since you set $\quad u_p=C_3x\sin(\pi x)+C_4x\cos(\pi x)\quad$ one can suppose that the ODE isn't $\quad y''+a^2y=\sin(\pi x)\quad$ but is $\quad y''+\pi^2y=\sin(\pi x)$. Introducing in the ODE the symbol $a$ which is a known constant $a=\pi$ is the cause of confusion in the search for $y_p$.


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.