y' - y = x^2 y' - y/x = x 

Use substitution: u = y/x y = ux y' = xu' + u 

xu' + u - y/x = x xu' + u - u = x xu' = x u' = 1 

u = ∫ 1 dx u = x + C y/x = x + C y = x^2 + Cx 

NOTE: We cannot find for sure that y = x^2 + 3x unless we have some initial condition, such as y(1) = 4Mathmom · 4 years ago1Thumbs up 0Thumbs downCommentReport Abuse

xy' - y = x² 

Dividing both sides by x. 

y' - (y/x) = x 

Now this becomes a first-order linear D.E. 

Integrating factor = e^(∫-1/x dx) = 1/x 

Multiplying both sides by the integrating factor: 

y'/x - (y/x²) = 1 

d/dx[y/x] = 1 

Integrating both sides: 

y/x = x + C 

y = x² + Cx 

Now you must've had some initial value along with this problem, since you said that the answer is y = 3x + x². This means that C = 3.