# solution to a cauchy euler form 2nd order

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.

Given:

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

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.