General solution of a second order linear ODE.

Question

$\DeclareMathOperator{\d}{\mathrm{d}}$I want to find the general solution of $$\frac{\d^2 u(t)}{\d t^2}-4t\frac{\d u(t)}{\d t}+(4t^2-2) u(t) = 0$$ I know that $u(t) = A\cdot e^{t^2}$ is a solution, but I'm stuck on finding this $A$. Any hints or suggestions?

Answer 1

$$u''-4t u'+(4t^2-2) u = 0 \tag 1$$ You know a solution $u(t) = A\cdot e^{t^2}$. Then consider that $A$ is no longer a constant, but an unknown function $A(t)$. $$u(t)=A(t)e^{t^2} \tag 2$$ $$u'=e^{t^2}(A'+2tA)$$ $$u''=e^{t^2}(A''+4tA'+2A+4t^2A)$$ $$e^{t^2}(A''+4tA'+2A+4t^2A)-4t(e^{t^2}(A'+2tA))+(4t^2-2)A e^{t^2} = 0$$ After simplification the equation is reduced to : $$e^{t^2}A''=0$$ Thus $A''=0$ which leads to $$A=c_1t+c_2$$ Bringing it back into Eq.$(2)$ : $$u(t)=(c_1t+c_2)e^{t^2}$$