- #1

- 109

- 2

## Main Question or Discussion Point

I've come across this problem while self studying Ordinary Differential Equations and I really need help. The problem asks me to simply eliminate derivatives, I do not need to separate. The book shows the answer, but not the steps.

problem:

[itex]\frac{dy}{dx}[/itex]= [itex]\frac{y^2}{1-xy}[/itex]

answer:

ln(y)=xy

So far, I'm here...

[itex]\frac{dy}{dx}[/itex]= [itex]\frac{y^2}{1-xy}[/itex]

y' = [itex]\frac{y^2}{1-xy}[/itex]

y' = [itex]\frac{y}{\frac{1}{y}-x}[/itex]

y = y'([itex]\frac{1}{y}[/itex]-x)

y = [itex]\frac{y'}{y}[/itex] - y'x

[itex]\frac{y'}{y}[/itex] = y+ y'x

now integrating both sides we get..

ln (y) = ∫(y + y'x dx)

But how do I integrate y+y'x?

problem:

[itex]\frac{dy}{dx}[/itex]= [itex]\frac{y^2}{1-xy}[/itex]

answer:

ln(y)=xy

So far, I'm here...

[itex]\frac{dy}{dx}[/itex]= [itex]\frac{y^2}{1-xy}[/itex]

y' = [itex]\frac{y^2}{1-xy}[/itex]

y' = [itex]\frac{y}{\frac{1}{y}-x}[/itex]

y = y'([itex]\frac{1}{y}[/itex]-x)

y = [itex]\frac{y'}{y}[/itex] - y'x

[itex]\frac{y'}{y}[/itex] = y+ y'x

now integrating both sides we get..

ln (y) = ∫(y + y'x dx)

But how do I integrate y+y'x?