How to factorise the following linear differential equation?

Question

I have been doing factorisation problems on linear differential equation and I faced a problem which I could not solve: $(D^2-(3/x )D+3/x^2)y=e^x+3$ How to break the differential expression into linear factors? Is there any general technique or rule of hand to write the factors at a glance?

Answer 1

Rewriting it as:$$x^2y''-3xy'+3y=x^2(e^x+3)$$ This is Euler's equation: $$(xD(xD-1)-3xD+3)y=x^2(e^x+3)$$ $$(xD(xD-1)-3(xD-1)y=x^2(e^x+3)$$ $$(xD-1)(xD-3)y=x^2(e^x+3)$$

Where the operator $D=\frac{d} {dx}$ And $$xD(xD-1)y=xDxD y-xDy$$ $$xD(xD-1)=x^2y''+xy'-xy'=x^2y''$$

Answer 2

This is an Euler Cauchy Equation. For the homogeous part, try plugging in $y=x^k$. You will get to solve $k(k-1)-3k+3=0$, so $k=1$ or $k=3$. Hence two linearly independent solutions are given by $x$ and $x^3$. The equation is not an equation where you can use the characteristic equation since the coefficients in front of $D=d/dx$ are not constant but depend on $x$.