Differential equation variation of parametres

Question

Find the general solution of

$y'' + \dfrac{7}{x} y' + \dfrac{8}{x^2} y = 1, x > 0$

I don't even know how to solve the homogeneous version because it involves variables...

Does anyone know how to solve it?

Answer 1

It is of Euler differential equation type. Here is a related problem. You should have the following solution

$$ y(x) ={\frac {{\it c_2}}{{x}^{4}}}+{\frac {{\it c_1}}{{x}^{2}}}+\frac{{x}^{2}}{24}.$$

Answer 2

$y=x^n$, $y'=nx^{n-1}$, $y''=n(n-1)x^{n-2}$. So $x^2y''+7xy'+8y=x^n[n(n-1)+7n+8]=0$ solve this equation for $n$.

So that gives $n^2+6n+8=0$ or $n=-4,-2$. The solution of homogeneous eq is $y_h=c_1x^{-4}+c_2x^{-2}$.

The particular solution here can be obtained by undetermined coefficient and guessing form of solution $y_p=Ax^2$. use that to solve $x^2y''+7xy'+8y=x^2$ here we get $A(2(2-1)+7*2+8)=1$ or $A=1/24$.

Now general solution is $y_g=y_h+y_p=c_1x^{-4}+c_2x^{-2}+x^2/24$.