# Find general solution to 2nd order differential equation

Given $$\frac{d^2u}{dt^2}-4t\frac{du}{dt}+(4t^2-2)u=0$$ Find the general solution.

I believe I'm supposed to find the fundamental solutions first, by rewriting the equation into two $1^{st}$ order differential equations, and I have done so. However, this will lead to pretty ugly eigenvalues ($\lambda_{1,2}=2t\pm\sqrt{8t^2-2}$ and therefore ugly eigenvectors. Am I supposed to do it this way or is there an easier/quicker method?

Update Using Wolframalpha I've managed to find a solution: $$y(t)=c_1e^{t^2}+c_2te^{t^2}$$ However, I would like to know why this is so.

Apply the method in http://eqworld.ipmnet.ru/en/solutions/ode/ode0205.pdf:

Let $u=e^{t^2}v$ ,

Then $\dfrac{du}{dt}=e^{t^2}\dfrac{dv}{dt}+2te^{t^2}v$

$\dfrac{d^2u}{dt^2}=e^{t^2}\dfrac{d^2v}{dt^2}+2te^{t^2}\dfrac{dv}{dt}+2te^{t^2}\dfrac{dv}{dt}+(4t^2+2)e^{t^2}v=e^{t^2}\dfrac{d^2v}{dt^2}+4te^{t^2}\dfrac{dv}{dt}+(4t^2+2)e^{t^2}v$

$\therefore e^{t^2}\dfrac{d^2v}{dt^2}+4te^{t^2}\dfrac{dv}{dt}+(4t^2+2)e^{t^2}v-4t\left(e^{t^2}\dfrac{dv}{dt}+2te^{t^2}v\right)+(4t^2-2)e^{t^2}v=0$

$e^{t^2}\dfrac{d^2v}{dt^2}+4te^{t^2}\dfrac{dv}{dt}+(4t^2+2)e^{t^2}v-\left(4te^{t^2}\dfrac{dv}{dt}+8t^2e^{t^2}v\right)+(4t^2-2)e^{t^2}v=0$

$e^{t^2}\dfrac{d^2v}{dt^2}=0$

$\dfrac{d^2v}{dt^2}=0$

$v=c_1+c_2t$

$u=c_1e^{t^2}+c_2te^{t^2}$

• Is the use of $u=e^{t^2}v$ an educated guess or is there some sort of theory that describes why this is 'the way to go'? Apr 2, 2016 at 18:17