I just started learning about differential equations and encountered following equation:
$$ y'y = y +1 $$
Wolfram alpha provided the following explanation:
here
But I'm not sure how the integration is performed. What rules are used?
How does $\int{\frac{y'y}{y+1}dx}$ become $-\log(y + 1) + y$?
Best Answer
They use substitution, namely that $y'dx = dy$, so you can rewrite $$ \int\frac{y'y}{y+1}dx = \int\frac{y}{y+1} dy = \int\left(\frac{-1}{y+1} + 1\right)dy $$ Using this substitution to solve a differential equation is called "solution by separation", and an equation where this is possible to do is called a "separable equation". The general requirement for a separable equation is that you can get the equation to the form $f(y)y' = g(x)$, which then becomes $\int f(y) dy = \int g(x)dx$.