Question

(1-x^2) dy/dx + xy = ax​

Answer 1

Step-by-step explanation:(1-x^2)y' + xy = ax.  

y' + x/(1-x^2)y = ax/(1-x^2).  

integrating factor: e^(∫x/(1-x^2)dx) = e^(-1/2ln(1-x^2)) = 1/√(1-x^2).  

becomes:  

d/dx[y/√(1-x^2)] = ax/(1-x^2) * (1/√(1-x^2))  

d/dx[y/√(1-x^2)] = ax/(1-x^2)^(3/2).  

y/√(1-x^2) = ∫ax/(1-x^2)^(3/2) dx.  

I think you can take it from here. just a u-sub for the integral, and don't forget about the constant.