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I've got an equation

$$e^x(y^2-4y' -4y)=4y'$$

after some transformations, i recieve

$$y= \frac{-4}{e^x+C_2(x)}$$

$$y'=\frac{4(e^x+C'_2(x))}{(e^x+C_2(x))^2}$$ but when i add these equations to the first one $C_2(x)$ sill remains in equation.. Can somebody sole it?

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  • $\begingroup$ What does $C_2(x)$ represent? Should that just be an arbitrary constant, not an arbitrary function? $\endgroup$ – tilper Apr 24 '17 at 15:57
  • $\begingroup$ It is a Riccati type. $\endgroup$ – hamam_Abdallah Apr 24 '17 at 15:59
  • $\begingroup$ Arbitrary constant, i should have be more precise! sorry! $\endgroup$ – Mikkey Apr 24 '17 at 15:59
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    $\begingroup$ @Mikkey: This is a Separable Equation and you would have $$\displaystyle \int \dfrac{4}{y(y-4)}~ dy = \int \dfrac{e^x}{e^x + 1}~dx$$ $\endgroup$ – Moo Apr 24 '17 at 16:05
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    $\begingroup$ @Mikkey, if $C_2$ is just a constant then you should say $C_2$ and not $C_2(x)$. Also, since $C_2$ is a constant then $C_2' = 0$. $\endgroup$ – tilper Apr 24 '17 at 16:12
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As per the comment discussion on the question, $C_2(x)$ should really be $C_2$ since $C_2$ is just a constant and does not depend on $x$.

You found $y(x) = \dfrac{-4}{e^x + C_2}$.

So then $y'(x) = \dfrac{(e^x + C_2)(0) - (-4)(e^x)}{(e^x + C_2)^2} = \dfrac{4e^x}{(e^x+C_2)^2}$

When you plug this into the LHS of the original equation, you get:

\begin{align*} e^x(y^2 - 4y' - 4y) &= e^x \left(\frac{16}{(e^x+C_2)^2} - \frac{16e^x}{(e^x+C_2)^2} + \frac{16}{e^x+C_2}\right)\\[0.3cm] &= e^x \left(\frac{16}{(e^x+C_2)^2} - \frac{16e^x}{(e^x+C_2)^2} + \frac{16(e^x+C_2)}{(e^x+C_2)^2}\right)\\[0.3cm] &= e^x \left(\frac{16}{(e^x+C_2)^2} - \frac{16e^x}{(e^x+C_2)^2} + \frac{16e^x}{(e^x+C_2)^2} + \frac{16C_2}{(e^x+C_2)^2}\right)\\[0.3cm] &= e^x \left(\frac{16}{(e^x+C_2)^2} + \frac{16C_2}{(e^x+C_2)^2}\right)\\[0.3cm] &= \frac{16e^x(1+C_2)}{(e^x+C_2)^2} \end{align*}

But in the RHS, you get: $$ 4y' = \frac{16e^x}{(e^x+C_2)^2} $$

So very close, but these are not quite the same. So your $y(x)$ isn't correct.

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