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Question Number 50431 by Abdo msup. last updated on 16/Dec/18

$$solve{xy}^{{}^{\prime }}+y=\frac{2x}{\sqrt{1-x^4}}$$

Commented by Abdo msup. last updated on 16/Dec/18

$$\left(he\right){xy}^{{}^{\prime }}+y=0\Rightarrow {xy}^{{}^{\prime }}=-y\Rightarrow \frac{y^{{}^{\prime }}}{y}=-\frac{1}{x}\Rightarrow ln\mid y\mid =-ln\mid x\mid +k\Rightarrow$$$$y\left(x\right)=\frac{K}{\mid x\mid }lettakex>0\Rightarrow y=\frac{K}{x}mvcmethodgive$$$$y^{{}^{\prime }}=\frac{K^{{}^{\prime }}}{x}-\frac{K}{x^2}and\left(e\right)\iff K^{{}^{\prime }}-\frac{K}{x}+\frac{K}{x}=\frac{2x}{\sqrt{1-x^4}}\Rightarrow$$$$K^{{}^{\prime }}=\frac{2x}{\sqrt{1-x^4}}\Rightarrow K\left(x\right)=\int \frac{2xdx}{\sqrt{1-x^4}}changement{x}^2=sint$$$$giveK\left(x\right)=\int \frac{costdt}{\sqrt{1-{sin}^{2}t}}=\int dt+c=t+\lambda =arcsin\left(x^2\right)+\lambda$$$$\Rightarrow y\left(x\right)=\frac{arcsin\left(x^2\right)+\lambda }{x}=\frac{\lambda }{x}+\frac{arcsin\left(x^2\right)}{x}.$$$$$$$$$$

Answered by tanmay.chaudhury50@gmail.com last updated on 16/Dec/18

$$xdy+ydx=\frac{2x}{\sqrt{1-{\left(x^2\right)}^2}}dx$$$$d\left(xy\right)=\frac{d\left(x^2\right)}{\sqrt{1-{\left(x^2\right)}^2}}$$$$\int d\left(xy\right)=\int \frac{d\left(x^2\right)}{\sqrt{1-{\left(x^2\right)}^2}}$$$$xy={sin}^{-1}\left(x^2\right)+c$$$$$$

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