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$u(x)=y+x$, solve $\frac{dy}{dx} = (y+x)^2$

I'm unsure about what dy/dx becomes.


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If $u=y+x$, then ${du\over dx}={dy\over dx}+1$.

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what if $ u(x) = y^3 $. what would be $ \frac{dy}{dx} $ in this case. – 40Plot Jan 25 '13 at 4:09
Hint: $(y^2)'=(y\cdot y)'=2yy'$ and $(y^3)'=(y^2\cdot y)'$ – Ian Mateus Jan 25 '13 at 4:16
40Plot, it's just the chain rule. If $u(x)=y^3$, then $du/dx=3y^2(dy/dx)$, so $dy/dx=(3y^2)^{-1}(du/dx)=(3u^{2/3})^{-1}(du/dx)$ – Gerry Myerson Jan 25 '13 at 5:37

If $u(x)=x+y$ then $u'(x)=x'+y',~~ y=y(x)$ so $u'(x)=1+y'$ so your OE became $$u'-1=y'=u^2$$ Now solve $$u'=1+u^2$$ or $$\frac{du(x)}{1+u^2}=dx$$

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thank you! I see i now – 40Plot Jan 25 '13 at 4:01

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