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The O. D. E. I am trying to solve is $$y''(-y')^m=-y^ny',\quad y(1)=1,\quad y'(1)=-1,\quad (*)$$ where $m$ and $n$ are integers greater than one. If we observe that $$y''(-y')^m=-\frac{1}{m+1}[(-y)^{m+1}]',$$ and that $$y^ny'=\frac{1}{n+1}[y^{n+1}]',$$ we can integrate the equation ($*$) and obtain the following equation: $-\frac{1}{m+1}(-y')^{m+1}+\frac{1}{n+1}(y)^{n+1}=C$ where $C$ is a constant. From now on, I don't know what to do.

Could someone give me an idea to complete the solution above or show me another way to find $y$ in the equation ($*$)?

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Methinks you are missing a $'$ in the RHS of your original equation (*) – Jaime Oct 2 '12 at 20:29
Thank you Jaime. – Son Oct 2 '12 at 20:33

Hint: Since $$ y''(-y')^m=\frac{(-1)^m}{m+1}[(y')^{m+1}]'=-y^ny'=-\frac{1}{n+1}(y^{n+1})' $$ we have $$ \frac{(-1)^m}{m+1}(y')^{m+1}+\frac{1}{n+1}y^{n+1}=C, $$ and taking into account the initial conditions $y(1)=1=-y'(1)$, we find that $$ C=\frac{1}{n+1}-\frac{1}{m+1}=\frac{m-n}{(n+1)(m+1)}. $$ So now we have $$ (y')^{m+1}=(-1)^m\frac{m+1}{n+1}\left(\frac{m-n}{m+1}-y^{n+1}\right), $$ i.e. $$ y'=\begin{cases} a_{m,n}\left(\frac{m-n}{m+1}-y^{n+1}\right)^{\frac{1}{m+1}} & m\ \text{ even }\\ a_{m,n}\left(y^{n+1}-\frac{m-n}{m+1}\right)^{\frac{1}{m+1}} & m\ \text{ odd } \end{cases}, $$ where $$\tag{1} a_{m,n}=\left(\frac{m+1}{n+1}\right)^{\frac{1}{m+1}}. $$ Hence \begin{eqnarray} \int_1^{y(t)}\left[\frac{m-n}{m+1}-u^{n+1}\right]^{-\frac{1}{m+1}}du&=&a_{m,n}(t-1)\quad m\ \text{ even }\tag{2}\\ \int_1^{y(t)}\left[u^{n+1}-\frac{m-n}{m+1}\right]^{-\frac{1}{m+1}}du&=&\pm a_{m,n}(t-1)\quad m\ \text{ odd }\tag{3} \end{eqnarray}

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