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Below is my solution to a differential equation. The answer that I got was different than the book's answer so I am assuming mine is wrong. I am hoping somebody can tell me where I went wrong.
Problem:
Solve the following differential equation. \begin{eqnarray*} \frac{dy}{dx} + y &=& xy^3 \\ \end{eqnarray*}
Answer:
Observe that this is a Bernoulli equation. \begin{eqnarray*} y^{-3} \frac{dy}{dx} + y^{-2} &=& x \\ z &=& y^{-2} \\ \frac{dz}{dz} &=& -2y^{-3 } \frac{dy}{dx} \\ \frac{1}{2} \frac{dz}{dx} + z &=& x \\ \frac{dz}{dx} + 2z &=& 2x \\ \end{eqnarray*} Now we have a linear differential equation so we look to find an integrating factor, $I$. \begin{eqnarray*} I &=& e ^{ \int 2 dx } = e^ {2x} \\ e^ {2x} \frac{dz}{dx} + 2e^ {2x}z &=& 2xe^{2x} \\ D( e{2x}z ) &=& 2xe^{2x} \\ e^{2x}z &=& \int 2xe^{2x} \\ \end{eqnarray*} Now to evaluate this integral, we use integration by parts with $u = x$, $dv = 2e^{2x}dx$ and $v = e^{2x}$. \begin{eqnarray*} \int 2xe^{2x} &=& xe^{2x} - \int e^{2x} \,\, dx = xe^{2x} - \frac{ e^{2x} }{2} + C_1 \\ e^{2x}z &=& xe^{2x} - \frac{ e^{2x} }{2} + C_1 \\ \frac{e^{2x}}{y^2} &=& xe^{2x} - \frac{ e^{2x} }{2} + C_1 \\ \frac{2}{y^2} &=& 2x - 1 + C e^{-2x} \\ \end{eqnarray*} However, the book gets: \begin{eqnarray*} \frac{2}{y^2} &=& 2x - 1 + C e^{2x} \\ \end{eqnarray*}

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    $\begingroup$ Shouldn't it be $-\frac12\frac{dz}{dx}+z=x$? $\endgroup$ – Akiva Weinberger Nov 1 '17 at 0:58
  • $\begingroup$ @Akiva Weinber, yes it should. $\endgroup$ – Bob Nov 1 '17 at 0:59
  • $\begingroup$ @Bob so you will get the answer. There is no reason for you to keep the question open! Your choice it is. $\endgroup$ – samjoe Nov 1 '17 at 1:53
  • $\begingroup$ How do I close hit? If I hit close what reason do I give? $\endgroup$ – Bob Nov 1 '17 at 22:32
  • $\begingroup$ @Akivia Weinber I would like to close this question and give you credit. Would you like to turn your message into an answer. $\endgroup$ – Bob Nov 1 '17 at 22:58
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Akiva Weinberger answered the question.

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