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Find a particular solution of $y''-2y=3-\frac{1}{t^2}$. The solutions of the corresponding homogeneous equation are $y_1=t^2$ and $y_2=t^{-1}$

Attempt: I am trying to find the particular solution by solving the following system: $$\begin{cases} u_1'y_1+u_2'y_2=0 \\ u_1'y_1'+u_2'y_2'=3-\frac{1}{t^2}\end{cases}$$

Here is the process:$$\begin{cases} u_1't^2+u_2't^{-1}=0 \\ 2u_1't+u_2't^{-2}=3-\frac{1}{t^2}\end{cases}$$

$$\begin{cases} u_1'=-u_2't^{-3} \\ -2u_2't^{-2}-u_2't^{-2}=3-\frac{1}{t^2} \end{cases}$$

$$\begin{cases} u_2'=-t^2+3 \\ u_1'=\frac{1}{t}-\frac{3}{t^3} \end{cases}$$

After integrating and writing down the particular solution I get: $y_p=t^2\ln t - \frac{1}{3} t^2 +3.5$. However, my book says the solution is $y_p=t^2\ln t +0.5$. I checked everything several times, but could not find where I made a mistake. Help please.

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  • $\begingroup$ None of your solutions satisfy ODE. $\endgroup$ – Kaster Mar 7 '13 at 6:32
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Acccording to the method, we should solve the following OE: $$u'_1=\frac{-y_2f(x)}W,~~~u'_2=\frac{y_1f(x)}W$$ where in $f(x)=3-t^{-2}$ and $W=W(y_1,y_2)=-3$ is the Wronskian of two solutions $y_1,y_2$. We have, then, $$u'_1=\frac{-t^{-1}(3-t^{-2})}{-3}\to u'_1=t^{-1}-\frac{1}{3t^3}\to u_1=\ln(t)+\frac{1}{6t^2}$$ and $$u'_2=\frac{t^2(3-t^{-2})}{-3}\to u'_2=-t^2+\frac{1}3\to u_2=(-t^3/3)+\frac{t}{3}$$

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  • $\begingroup$ Very nice! You "killed" the problem! +1 $\endgroup$ – Namaste Mar 7 '13 at 6:42

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