- #1
prace
- 102
- 0
Problem: [tex](y-yx^2)\frac{dy}{dx} = (y+1)^2[/tex]
So, the first thing I tried was just dividing the whole equation by [tex](y-yx^2)[/tex] and then factored out the y to get [tex]\frac{dy}{dx} = \frac{(y+1)^2}{y(1-x^2)}[/tex]. Next I expanded the numerator on the right side of the equation and then split them all into idividual fractions to get:
[tex]\frac{dy}{dx} = \frac{y}{1-x^2}+\frac{2}{1-x^2}+\frac{1}{y(1-x^2)}[/tex].
I just don't see how I can go from here?
So, the first thing I tried was just dividing the whole equation by [tex](y-yx^2)[/tex] and then factored out the y to get [tex]\frac{dy}{dx} = \frac{(y+1)^2}{y(1-x^2)}[/tex]. Next I expanded the numerator on the right side of the equation and then split them all into idividual fractions to get:
[tex]\frac{dy}{dx} = \frac{y}{1-x^2}+\frac{2}{1-x^2}+\frac{1}{y(1-x^2)}[/tex].
I just don't see how I can go from here?