The original equation is $t^2y''+3ty' +y=1/t,\ \ t>0.$

I have already solved for the complimentary assuming soln is $y=t^m$, so $y_1=t^{-1}$.

From there I have found a solution $y_2$ using reduction of order:
So $y_2 = vt^{-1}$.
Finding the derivatives and plugging into equation I finally get:
$y_2 = C_1t^{-1}\ln t + C_22t^{-1}$.

From here on I don't know what to do in order to solve this equation without using a wronskian formula.

Any help would be appreciated. Thank you.



$$\tag 1 t^2y''+3ty' +y= \dfrac{1}{t}, ~ t \gt 0$$

When we solve for the homogeneous equation, we get:

  • $y = t^m$
  • $y' = m t^{m-1}$
  • $y'' = m(m-1)t^{m-2}$

Substituting into the homogeneous part of $(1)$, we get:

$$(m+1)^2 t^m = 0 \rightarrow m_{1,2} = -1$$

This provides us with two (see case 2) solutions as:

$$y_h(t) = c_1\dfrac{1}{t} + c_2\dfrac{\ln t}{t}$$

Next, you can use Variation of Parameters or an Exact equation to solve for the particular solution (both are a bit messy). I prefer VoP.

This gives us a particular solution of:

$$y_p(t) = \dfrac{\ln^2 t}{2t}$$

Our final solution is:

$$y(t) = y_h(t) + y_p(t) = c_1\dfrac{1}{t} + c_2\dfrac{\ln t}{t}+ \dfrac{\ln^2 t}{2t}$$


The solution of your ODE is in fact $y(t) = C_1/ t + C_2\log (t)/ t + \log^2(t)/ (2t)$. Apparently you missed one solution. I hope and wish that this could help you.


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.