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.

  • $\begingroup$ None of your solutions satisfy ODE. $\endgroup$ – Kaster Mar 7 '13 at 6:32

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 $$u'_2=\frac{t^2(3-t^{-2})}{-3}\to u'_2=-t^2+\frac{1}3\to u_2=(-t^3/3)+\frac{t}{3}$$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.