Here is my proposed solution: \begin{align} x^2\frac{dy}{dx}&=\frac{4x^2-x-2}{\left(x+1\right)\left(y+1\right)}\tag{1}\\ \implies & \left(y+1\right)\frac{dy}{dx}=\frac{4x^2-x-2}{x^2\left(x+1\right)}\tag{2}\\ \implies & \int\frac{dy}{dx}\left(y+1\right)\:dx=\int\frac{4x^2-x-2}{x^2\left(x+1\right)}\:dx\tag{3}\\ \implies & \frac{y^2}{2}+y=\int\alpha\left(x\right)\:dx,\tag{4} \end{align} and here I'm letting $\alpha\left(x\right)$ denote the partial fraction decomposition of the integrand in the RHS, which is \begin{align} \alpha\left(x\right)=\frac{4x^2-x-2}{x^2\left(x+1\right)}=\frac{1}{x}-\frac{2}{x^2}+\frac{3}{x+1},\tag{5} \end{align} and this gives me \begin{align} \frac{y^2}{2}+y=\log\left|x\right|+\frac{2}{x}+3\log\left|x+1\right|+C.\tag{6} \end{align} Okay, so no problems there. But now the IC is such that $y\left(1\right)=1.$ I don't know what to do with these two numbers because this is an implicit answer and I cannot separate the $y$'s.

Thank you for your time,


1 Answer 1


$y(1)=1$ implies (by plugging 1 into both $x$ and $y$)

$$\frac12+1=\log 1+\frac21+3\log 2+C.$$

Now, solve for $C$.

  • $\begingroup$ Okay, so I just plug them in as-is? Haha maybe this was too easy. I was probably just over-thinking it because I thought the whole RHS had to be equal to 1 as $y\left(1\right)=1$, but then I couldn't figure out what to do with the $y$'s. Oh well. Thank you for answering! $\endgroup$
    – bjd2385
    Jan 30, 2015 at 18:57
  • 1
    $\begingroup$ Yep, you were definitely over thinking. $\endgroup$ Jan 30, 2015 at 18:58

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