Separation of variables. Differential equation.

Question

solomonzelman · Answer

$\large\color{black}{ \displaystyle xe^ydy+\frac{x^2+1}{y}dx=0 }$

$\large\color{black}{ \displaystyle xe^ydy=-\frac{x^2+1}{y}dx }$

$\large\color{black}{ \displaystyle xye^ydy=-(x^2+1)dx }$

$\large\color{black}{ \displaystyle ye^ydy=\frac{-(x^2+1)}{x}dx }$

$\large\color{black}{ \displaystyle ye^y\frac{dy}{dx}=\frac{-(x^2+1)}{x} }$

and integrate with respect to x on both sides (like last time)

solomonzelman · Answer

I just don't feel that you can add an $\large\color{black}{ \displaystyle \int }$ sign just like that you know, that is why I bring dx on the other side and integrate with respect to x.

solomonzelman · Answer

(same way I can instead of dividing by dx on both sides, divide by dy on both sides and then integrate with respect to y)

solomonzelman · Answer

just a little playing around algebraically.

anonymous · Answer

Oh wait. Why is it (y-1)^ey?

anonymous · Answer

I thought in integrating ye^y, Ill just use the idea of udv+vdu..

anonymous · Answer

Thank you so much @SolomonZelman =)

solomonzelman · Answer

yw