1. ## Second Order DE

I'm having trouble trying to solve the differential equation $y^{(2)} + (\frac{1}{4t^2})y=f cos(t), t>0$ given that $y_{1}=\sqrt{t}$ is a solution of the homogeneous equation. I have been trying to solve this by reduction of order and am having no luck. I beleive the correct answer to be $y(t)=\sqrt{t}[c_{1}+c_{2}ln(t)+\int_0^t f \sqrt{s} cos(s)[ln(t)-ln(s)]ds]$ but can't seem to figure it out. Any help would be appreciated.

2. Reduction of order certainly should work. Exactly what have you tried?

If $y(t)= u(t)t^{1/2}$ then $y'= u't^{1/2}+ (1/2)ut^{-1/2}$ and $y"= u"t^{1/2}+ u't^{-1/2}- (1/4)ut^{-3/2}$.

Putting that into the equation, $y"+ (1/(4t^2)y= u"t^{1/2}+ u't^{-1/2}- (1/4)ut^{-3/2}+ (1/4)ut^{-3/2}$[tex]= u"t^{1/2}+ u't^{-1/2}= f cos(t)[tex]. Let v= u' and that is ][tex]v't^{1/2}+ vt^{-1/2}= f cos(t)[tex], a linear first order equation. You might have to leave part of the solution in terms of an integral.

3. I have everything you have up to that point. That was the easy part. Solving for $u$ seems to be the problem. If I solve using $y_{2}=y_{1}(t)\int \frac{exp[-\int P(t)dt]}{y_1^2(t)}dt$ I get a solution, but not the given solution. If I try to solve for $u$ use using brute force method I get hung up and am unable to produce the given solution.

4. Why are you integrating it from zero? Looks to me it should be solved for $t_0>0$. Letting $y=u\sqrt{t}$ as HallsofIvy suggested, suppose I do it from $t_0$ and not zero, I get the expression:

$\int_{t_0}^{t}du=\int_{t_0}^{t}\left[\frac{c_1}{t}+\frac{1}{t}\int_{t_0}^t s^{1/2}f\cos(s)ds\right]dt$

Doing the left side and changing the variable names on the right so that we can follow it better:

$u(t)-u(t_0)=\int_{t_0}^{t}\left[\frac{c_1}{w}+\frac{1}{w}\int_{t_0}^w s^{1/2}f\cos(s)ds\right]dw$

$u(t)-u(t_0)=c_1(\ln(t)-\ln(t_0))+\int_{t_0}^{t}\int_{t_0}^{w} \frac{1}{w}s^{1/2}f\cos(s)ds dw$

Now, can you justify switching the order of integration in order to arrive at a solution which looks like yours but starts at $t_0$?

5. I see what you did there. Thank you for your help, both of you.