Solve the Cauchy Euler Equations. $x^2 y''+7x y'+13y=\ln⁡ x$

Question

I am trying to solve this question but dont know how to start and what is the method to follow Solve the Cauchy Euler Equations.

$$x^2 \frac {d^2 y}{dx^2 }+7x \frac{dy}{dx}+13y=\ln ⁡x$$

Answer 1

$$x^2 y''+7x y'+13y=\ln⁡ x$$ Substitute $x=e^z \implies z=\ln x$

Then $$y'=\frac {dy}{dx}=\frac {dy}{dz}\frac {dz}{dx}=\frac 1 x\frac {dy}{dz}$$ $$y''=\frac {d^2y}{dx^2}=-\frac 1 {x^2}\frac {dy}{dz}+\frac 1 {x^2}\frac {d^2y}{dz^2}$$ The equation becomes $$y''+6y'+13y=z$$

can you take it from there ?

Answer 2

Hint:

Assume $$u=x^{\lambda}$$ $$u^{\prime}=\lambda x^{\lambda-1}$$ $$u^{\prime\prime}=\lambda(\lambda-1)x^{\lambda-2}$$