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I'm stuck on this differential equation:

$$y''+9y=36x\cos(3x), \quad \text{with }y(0)=-3, y'(0)=4$$

I know the homogenous equation is: $y_H(x)=A\cos(3x)+B\sin(3x)$

Now to find the particular solution...

I believe this next step is right.

$$Z_P(x)=A(x)e^{3ix}$$

Then we find the second derivative.

$$Z''_P(x)=A''(x)e^{3ix}+6iA''(x)e^{3ix}-9A(x)e^{3ix}$$

After plugging $Z_P$ and $Z''_P$ into the equation we arrive at $A''+6iA=36x$

Where do we go now? I'm not sure what to do.

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  • $\begingroup$ Isn't $A=-6ix$ a solution? $\endgroup$
    – nbubis
    Nov 10, 2014 at 18:51
  • $\begingroup$ Umm... I'm not sure. How did you arrive at that answer? $\endgroup$
    – FofX
    Nov 10, 2014 at 18:55
  • $\begingroup$ Oh wait. Is the equation $y'' + y' = ...$ or $y'' + y = ...$? Those two equations are completely different. $\endgroup$
    – Simon S
    Nov 10, 2014 at 21:00
  • $\begingroup$ Sorry I made that really confusing. The equation is $y''+9y=36xcos(3x)$ $\endgroup$
    – FofX
    Nov 11, 2014 at 19:49

2 Answers 2

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I'm not sure what method you're using there.

If you want to use variation of parameters, then we use an Ansatz of a linear combination of the two homogeneous solutions $\cos(3x)$ and $\sin(3x)$, $$y_p(x) = a(x)\cos(3x) + b(x)\sin(3x)$$ Then after some algebra and imposing the condition $a'(x)\cos(3x) + b'(x)\sin(3x) = 0$, we have expressions for $a$ and $b$,

$$a'(x) = - \frac{\cos(3x).36x\cos(3x)}{W}, \ \ \ b'(x) = -\frac{\sin(3x).36x\cos(3x)}{W}$$

where $W$ is the Wronksian of the two homogenous solutions $\cos 3x$ and $\sin 3x$, $$W = \cos(3x)(\sin(3x))' - \sin(3x)(\cos(3x))' = 3\cos^2(3x) + 3\sin^2(3x) = 3$$

Hence to find $a$ and $b$, integrate the expresssions for $a'$ and $b'$. Finally, construct the particular solution $y_p$.

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  • $\begingroup$ I'm just in Calculus II so we're really just covering the basics. This looks a little too advanced for the stuff we've covered. I'm not sure what the method is called I was just going off of what my instructor's notes were. I do appreciate the help though. $\endgroup$
    – FofX
    Nov 10, 2014 at 19:18
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The corresponding homogeneous equation is $$y''+9y'=0\ ,$$ and its general real solution is $$y_h(x)=C_1+C_2e^{-9x}, \qquad C_1, C_2\in{\mathbb R}\ .\tag{1}$$ The right side of the given inhomogeneous equation is obviously not of the form $(1)$. Therefore the textbook "Ansatz" for a particular solution $x\mapsto y_p(x)$ is given by $$y_p(x):=(ax+b) \cos(3x)+(cx+d)\sin(3x)\ .$$ Now plug this in into the given ODE and compare coefficients for $\cos x$, $x\cos x$, $\sin x$, $x\sin x$. Solve the resulting system of linear equations for $a$, $b$, $c$, $d$.

The general solution of the given ODE is then given by $$y(x):=y_h(x)+y_p(x)=C_1+C_2e^{-9x}+(ax+b) \cos(3x)+(cx+d)\sin(3x)\ ,$$ where $a$, $b$, $c$, $d$ have known numerical values. Finally determine $C_1$, $C_2$ using the initial conditions.

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