ODE particular solution of the homogeneous equation

[Dragonfall]

$$y''(t)+A^2y(t)=f(t), t>0, y(0)=B, y'(0)=C, A, B, C\in\mathbb{R}$$

$$e^{iAt}$$ is a particular solution of the homogeneous equation. I can multiply it by some arbitrary function and find another solution of the homogeneous case, but when I try with the f(t) on the RHS, I can't do it. Anyone help?

 

[HallsofIvy]

I'm not at all sure what you mean by "I can multiply it by some arbitrary function and find another solution of the homogeneous case" Since the two solutions to the characteristic equation for this problem are Ai and -Ai, the two independent solutions to the homogeneous equation are
$e^{iAt}$ and $e^{-iAt}$. What do you mean by "try with the f(t) on the RHS"? Are you talking about "variation of parameters"?

 

[Dragonfall]

I meant by reduction of order.

 

[HallsofIvy]

Ah! Then you are not multiplying by "some arbitrary function". You are multiplying by an unknown function and then determining what that function must be in order to solve the equation.

Since you know that $e^{iAt}$ satisfies the homogenous equation, you look for a solution to the entire equation of the form [itex\y= u(t)e^{iAt}[/itex]. Then $y'= u'e^{iAt}+ iAu e^{iAt}$ and $y"= u"e^{iAt}+ 2iAu'e^{iAt}- A^2ue^{iAt}$. Putting those into the original equation, the $A^2ue^{iAt}$ terms cancel leaving $u"e^{iAt}+ 2iAu'e^{iAt}= f(t)$. Since there are only first and second derivatives of u in that, let v= u' and the differential equation reduces to first order: $v'e^{iAt}+ 2iAve^{iAt}= f(t)$ or $v'+ 2iAv= f(t)e^{-iAt}$
That's a linear first order differential equation so there is a formula for the integrating factor.