I have here this problem, where I want to determine the general solution:
$$ u'= - \frac{2v}{t^2}+ te^t $$ $$ v'=-u+t $$ $ t \in \mathbb{R}^+ $
My idea is here to use variation of parameters, but how do I do this with the system? Any help very appreciated!
Hint: $$v'=-u+t$$ Differentiate: $$v''=-u'+1$$ Substitute this in the first DE: $$u'= - \frac{2v}{t^2}+ te^t$$ $$-v''+1=- \frac{2v}{t^2}+ te^t$$ $$-t^2v''+{2v}=t^3e^t-t^2$$ This is Cauchy-Euler 's differential equation.