Find the general solution of the nonhomogeneous differential equation

$$y'' + 4y =\sin(\alpha t).$$

For all possible real $\alpha$ decide which values of $\alpha$ gives a limited amount of solutions.

I found the homogeneous solution which yields: $y_h= C_1\sin(2t) + C_2\cos(2t)$

Any tips/solutions?

  • $\begingroup$ What do you mean by limited amount of solutions? $\endgroup$
    – David H
    May 20 '14 at 12:15
  • $\begingroup$ doesnt this get infinite amount of solutions since I have alpha? or something? $\endgroup$
    – gger234
    May 20 '14 at 12:24
  • $\begingroup$ So now you need to find a particular solution of the inhomogeneous equation. This will be of the form $A\sin(at)+B\cos(at)$ where you work out the values of $A$ and $B$ by substituting into the equation, except that if $a=2$ then you have to use $t(A\sin(at)+B\cos(at))$. $\endgroup$ May 20 '14 at 12:50


$y=sin(\alpha t)$ is a solution of $$\alpha^2y+y''=0$$ $$(D^2+\alpha^2)y=0$$ so apply

$$(D^2+\alpha^2)(y''+4y)=(D^2+\alpha^2)(sin(\alpha t))$$ $$(D^2+\alpha^2)(D^2+4)(y)=0$$ $$(D^4+(\alpha^2+4)D^2+4\alpha^2)y=0 $$

Thus, we have a homogenius equation with characterestic polynomial $$\lambda^4+(\alpha^2+4)\lambda^2+4\alpha^2=0 $$

Note: I used the $D$ as derivative operator.


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