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I have the following ODE: $$y'' + 4y' + 3y = 65\cos(2x)$$ I first used the method of undetermined coefficients to find the solution to this equation, which was relatively easy, and found that the solution was. $$C_1e^{-3x}+C_2e^x+8\sin(2x)-\cos(2x)$$ where to the my $y_h =C_1e^{-3x}+C_2e^x$, I used this $y_h$ in the start of my problem of doing variation of parameters for the same ODE, using the Wronskian and the general formula for computing the particular solution, but ended up getting my $y_p = 0$, which can't be right since thats not what I got in my undetermined coefficients method. I have a feeling I'm going about this the wrong way. What is the correct way to compute this ODE using variation of parameters?

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  • $\begingroup$ I tried, whenever I ask a question and add tags, it doesn't let me do a lot of the tags related to the subject since my reputation points are not 1k yet. $\endgroup$ – Collaptic Nov 16 '15 at 5:04
  • $\begingroup$ $y_h$ should be in $x$, not $t$ $\endgroup$ – Dylan Nov 16 '15 at 15:47
  • $\begingroup$ Please show us what you've done (edit the post) so we can diagnose the problem $\endgroup$ – Dylan Nov 16 '15 at 15:48
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    $\begingroup$ Yes, this seems like a straightforward application of the formulas. Begin with the trial form $ y_p(t) = c_1(x)y_1(x) + c_2(x) y_2(x) $ where $y_1(x)=e^{-3x}$ and $y_2(x)=e^x$. Then the formulas are: $$ c_1(x) = -\int\frac{y_2(x)g(x)}{y_1(x)y_2'(x)-y_1'(x)y_2(x)} $$ $$ c_1(x) = \int\frac{y_1(x)g(x)}{y_1(x)y_2'(x)-y_1'(x)y_2(x)} $$ where $g(x)=65\cos(2x)$. You basically plug in for $y_1$, $y_2$, and $g$, then plug into the first formula, and should get the answer. Where is this going wrong? $\endgroup$ – rajb245 Nov 16 '15 at 16:06
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First of all, let me point out that you've made a mistake with your complementary solution $y_h$. It should be: $$y_h(x)=c_1 e^{-3x}+c_2 e^{\color{red}{-x}}$$ I will show this in the first part of my answer.


Let's start by finding the complementary solution. $$\frac{d^2 y}{dx^2}+4\frac{dy}{dx}+3y=0$$ We make the ansatz $y=e^{\lambda x}$ to obtain the characteristic equation: $$\lambda^2+4\lambda+3=0$$ $$(\lambda+1)(\lambda+3)=0$$ $$\lambda=-1,-3$$ This implies that the complementary solution is: $$y_h(x)=c_1 e^{-3x}+c_2 e^{-x}$$


The basis solutions are $e^{-3x}$ and $e^{-x}$. Thus, the Wronskian is: $$W(x)=\begin{vmatrix} e^{-3x} & e^{-x} \\ -3e^{-3x} & -e^{-x} \end{vmatrix}=2e^{-4x}$$ Let $f(x)=65\cos(2x)$. The particular solution will thus be given by: $$y_p(x)=v_1(x) e^{-3x}+v_2(x) e^{-x}$$ Where: $$v_1(x)=-\int \frac{f(x)\cdot e^{-x}}{W(x)}~dx=-\int \frac{65\cos(2x)\cdot e^{-x}}{2e^{-4x}}~dx=-\frac{65}{2}\cdot \int e^{3x}\cos(2x)~dx$$ And: $$v_2(x)=\int \frac{f(x)\cdot e^{-3x}}{W(x)}~dx=\int \frac{65\cos(2x)\cdot e^{-3x}}{2e^{-4x}}~dx=\frac{65}{2}\cdot \int e^x\cos(2x)~dx$$ Can you continue?

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