Answer to Question #210410 in Differential Equations for Bilal

Question

xy"+2y'=0 reduced into y

Answer 1

Given xy"+2y'=0

let z=y′. ---------------------------------(i)

Then, dz/dx=y".

Given equation reduces to,

x(dz/dx)=−2z, ( by using separable variables )

=>−dz/z=2dx/x.

Integrating both sides, we get 

=>-logz =2logx+Log(c),where c is an arbitrary constant.

=>-logz =log(cx^2) 

=> z=Cx^(-2), where C is an arbitrary constant.

Now, from equation (i).

We get,

=>dy/dx=C x^(-2)

Integrate above , we get

=>y=-C(1/x)+c1