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Question Number 12814 by syambabu087@gmail.com last updated on 02/May/17

Commented by FilupS last updated on 02/May/17

$$x(y^n-y)+\frac{dy}{dx}=0$$$$\frac{dy}{dx}=-x(y^n-y)$$$$\frac{1}{y^n-y}\frac{dy}{dx}=-x$$$$\int \frac{1}{y^n-y}\frac{dy}{dx}dx=-\int xdx$$$$\int \frac{1}{y^n(1-y^{1-n})}dy=-\int xdx$$$$u=1-y^{1-n}\Rightarrow du=(n-1)y^{-n}dy$$$$\frac{1}{n-1}\int \frac{1}{u}du=-\int xdx$$$$\frac{1}{n-1}\ln \left(u\right)=-\frac{1}{2}{x}^2+C$$$$\frac{1}{n-1}\ln (1-y^{1-n})=-\frac{1}{2}{x}^2+C$$$$\ln (1-y^{1-n})=(1-n)\frac{1}{2}{x}^2+C$$$$1-y^{1-n}=e^{\frac{1}{2}(1-n)x^2}C{e}^{x+C}=e^{x}C$$$$y^{1-n}=1-e^{\frac{1}{2}(1-n)x^2}C$$$$y\left(x\right)={(1-e^{\frac{1}{2}(1-n)x^2}C)}^{\frac{1}{1-n}}$$$$$$$$y\left(0\right)=2$$$$\therefore {(1-e^{\frac{1}{2}(1-n)0^2}C)}^{\frac{1}{1-n}}=2$$$${(1-C)}^{\frac{1}{1-n}}=2$$$$1-C=2^{1-n}$$$$C=1-2^{1-n}$$$$$$$$\therefore y\left(x\right)={(1-(1-2^{1-n})e^{\frac{1}{2}(1-n)x^2})}^{\frac{1}{1-n}}$$

Commented by mrW1 last updated on 02/May/17

$$inyouransweryouhave$$$$x(y^n-y)+\frac{dy}{dx}=0$$$$$$$$butinthequestionitis$$$$x(y^n+y)+\frac{dy}{dx}=0$$

Answered by sma3l2996 last updated on 03/May/17

$$(y^n+y)x=-\frac{dy}{dx}\iff -xdx=\frac{dy}{y^n+y}$$$$\frac{-1}{2}{x}^2+c=\int \frac{dy}{y(y^{n-1}+1)}$$$$u=\frac{1}{y^{n-1}+1}\Rightarrow du=\frac{-(n-1)y^{n-2}}{{(1+y^{n-1})}^2}dy=-\frac{(n-1)y^{n-1}}{y{(y^{n-1}+1)}^2}dy$$$$y^{n-1}=-\frac{u-1}{u}$$$$du=\frac{(n-1)\times u}{y(y^{n-1}+1)}\times \frac{u-1}{u}dy=(n-1)\frac{u-1}{y(y^{n-1}+1)}dy$$$$\frac{dy}{y(y^{n-1}+1)}=\frac{du}{(n-1)(u-1)}$$$$\frac{-1}{2}{x}^2+c=\frac{1}{n-1}\int \frac{du}{u-1}=\frac{1}{n-1}ln(u-1)+a$$$$-\frac{1}{2}{x}^2+c_1=\frac{1}{n-1}ln(\frac{1}{y^{n-1}+1}-1)$$$$\frac{1}{y^{n-1}+1}-1=e^{-\frac{n-1}{2}{x}^2+(n-1)c_1}=C\times e^{-\frac{n-1}{2}{x}^2}$$$$(y^{n-1}+1)(C\times e^{-\frac{n-1}{2}{x}^2}+1)=1$$$$y^{n-1}=\frac{1}{C\times e^{-\frac{n-1}{2}{x}^2}+1}-1=-\frac{C\times e^{-\frac{n-1}{2}{x}^2}}{C\times e^{-\frac{n-1}{2}{x}^2}+1}$$$$y\left(0\right)=2$$$$-\frac{C}{C+1}=2^{n-1}\iff C=-\frac{2^{n-1}}{2^{n-1}+1}$$$$y^{n-1}=-\frac{C}{e^{\frac{n-1}{2}{x}^2}(C\times e^{-\frac{n-1}{2}{x}^2}+1)}$$$$=\frac{2^{n-1}}{(2^{n-1}+1)e^{\frac{n-1}{2}{x}^2}(\frac{-2^{n-1}}{(2^{n-1}+1)e^{\frac{n-1}{2}{x}^2}}+1)}=\frac{2^{n-1}}{(2^{n-1}+1)e^{\frac{n-1}{2}{x}^2}-2^{n-1}}$$$$y\left(x\right)=\frac{2}{{((2^{n-1}+1)e^{\frac{n-1}{2}{x}^2}-2^{n-1})}^{\frac{1}{n-1}}}$$

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