- #1
vanceEE
- 109
- 2
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?