How to find the general solutions of the following second-order ODE?

Question

I was wondering if there is a general solution of the following second-order homogeneous ODE.

$$ ty''(t)+(2-t)y'(t)=(1+\frac{\lambda}{t})y(t)\quad t>0 $$ where $\lambda$ is a positive constant.

Answer 1

$$ ty''(t)+(2-t)y'(t)=(1+\frac{\lambda}{t})y(t)\quad t>0 $$ HINT :

This looks like an ODE of the Bessel kind, but generalized. This draw us to try a change of function such as : $$y(t)=t^pe^{qt}z(t)$$ Compute $y'(t)$ and $y''(t)$. Put them into the ODE. Then determine the parameters $p,q,c$ such as the equation becomes on the form of a Bessel equation. Of course, this is an arduous calculus. This leads to : $$t^2z''+\left(2(p+1)t+(2q-1)t^2 \right)z'+\left((q-1)qt^2+(2q-p+2pq-1)t-\lambda+p(p+1) \right)z=0$$ To be checked. Nor quite sure of the calculus.

On can see that a simplification occurs with $$q=\frac12 \quad\text{and}\quad p=-\frac12$$ $$t^2z''+tz'-\frac14\left(t^2+4\lambda+1 \right)z=0$$ $$z(t)=c_1\text{I}_{\sqrt{\lambda+\frac14}}\left(\frac{t}{2}\right)+c_2\text{K}_{\sqrt{\lambda+\frac14}}\left(\frac{t}{2}\right)$$ I and K denote the modified Bessel functions of first and second kind. $$y(t)=c_1t^{-1/2}e^{t/2}\text{I}_{\sqrt{\lambda+\frac14}}\left(\frac{t}{2}\right)+c_2t^{-1/2}e^{t/2}\text{K}_{\sqrt{\lambda+\frac14}}\left(\frac{t}{2}\right)$$

Answer 2

Hint:

$ty''+(2-t)y'=\left(1+\dfrac{\lambda}{t}\right)y$

$t^2y''+(2t-t^2)y'-(t+\lambda)y=0$

Let $y=e^{at}u$ ,

Then $y'=e^{at}u'+ae^{at}u$

$y''=e^{at}u''+ae^{at}u'+ae^{at}u'+a^2e^{at}u=e^{at}u''+2ae^{at}u'+a^2e^{at}u$

$\therefore t^2(e^{at}u''+2ae^{at}u'+a^2e^{at}u)+(2t-t^2)(e^{at}u'+ae^{at}u)-(t+\lambda)e^{at}u=0$

$t^2u''+((2a-1)t^2+2t)u'+(a(a-1)t^2+(2a-1)t-\lambda)u=0$

Choose $2a-1=0$ , i.e. $a=\dfrac{1}{2}$ , the ODE becomes

$t^2u''+2tu'-\left(\dfrac{t^2}{4}+\lambda\right)u=0$