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Hi I'm trying to solve a separable DE with a worked solution.

Question: $y′ = (y−9x)^2$

Part of the Given Solution: Letting $v = y − 9x$ becomes $\frac{dv}{dx} = v(x)^2 − 9$

I get that $\frac{dv}{dx} = \frac{dy}{dx} - 9$ but how does the $\frac{dy}{dx}$ become a $v(x)^2$

Thanks

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Chris
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  • What did the original DE tell you? – colormegone Apr 04 '16 at 02:01
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    @Chris: Hint: write $y = v + 9x \implies y' = v' +9$. Now substitute those two relationships into the original DE, simplify and solve for $v(x)$, then substituent one more time for the final solution. – Moo Apr 04 '16 at 02:07
  • @Moo , thanks that was what I was looking for.. I knew it had to be something obvious I was missing! – Chris Apr 04 '16 at 02:30

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