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Question Number 103483 by bemath last updated on 15/Jul/20

$$y^2-u(u+y).\frac{dy}{du}=0$$

Answered by Dwaipayan Shikari last updated on 15/Jul/20

$$y^2=\frac{dy}{du}.u(u+y)$$$$y^2=\frac{dy}{du}vy(vy+y)\{u=vy,1=\frac{dv}{du}y+\frac{dy}{du}v$$$$1=\frac{dy}{du}v(v+1)$$$$1=(1-\frac{dv}{du}y)(v+1)$$$$1=v+1-\frac{dv}{du}y(v+1)$$$$\frac{v}{v+1}=\frac{dv}{du}\left(\frac{u}{v}\right)$$$$\int \frac{du}{u}=\int \frac{v+1}{v^2}dv$$$$logu=logv-\frac{1}{v}+C$$$$logy=-\frac{y}{u}+log\left(C_1\right)$$$$log\left(\frac{y}{C_1}\right)=-\frac{y}{u}$$$$y=C_1{e}^{\frac{-y}{u}}$$$$$$$$$$

Commented by Dwaipayan Shikari last updated on 15/Jul/20

$$IhavestartedmylessononDifferentialequations.Kindly$$$$checkmyanswer$$

Answered by bramlex last updated on 15/Jul/20

$$\frac{\mathrm{dy}}{\mathrm{du}}=\frac{{\mathrm{y}}^2}{{\mathrm{u}}^2+\mathrm{uy}}.\mathrm{set}\mathrm{y}=\mathrm{uw}$$$$\frac{\mathrm{dy}}{\mathrm{du}}=\mathrm{w}+\mathrm{u}\frac{\mathrm{dw}}{\mathrm{du}}$$$$\iff \mathrm{w}+\mathrm{u}\frac{\mathrm{dw}}{\mathrm{du}}=\frac{{\mathrm{u}}^2{\mathrm{w}}^2}{{\mathrm{u}}^2+{\mathrm{u}}^2\mathrm{w}}$$$$\mathrm{w}+\mathrm{u}\frac{\mathrm{dw}}{\mathrm{du}}=\frac{{\mathrm{w}}^2}{1+\mathrm{w}}$$$$\mathrm{u}\frac{\mathrm{dw}}{\mathrm{du}}=\frac{{\mathrm{w}}^2}{1+\mathrm{w}}-\mathrm{w}=\frac{-\mathrm{w}}{1+\mathrm{w}}$$$$\frac{1+\mathrm{w}}{\mathrm{w}}\mathrm{dw}=-\frac{\mathrm{du}}{\mathrm{u}}$$$$\int (\frac{1}{\mathrm{w}}+1)\mathrm{dw}=-\ln \mid \mathrm{u}\mid +\mathrm{C}$$$$\ln \mid \mathrm{w}\mid +\mathrm{w}=\mathrm{C}-\ln \mid \mathrm{u}\mid$$$$\ln \mid \mathrm{uw}\mid +\mathrm{w}=\mathrm{C}$$$$\ln \mid \mathrm{u}.\frac{\mathrm{y}}{\mathrm{u}}\mid +\frac{\mathrm{y}}{\mathrm{u}}=\mathrm{C}$$$$\mathrm{u}\ln \mid \mathrm{y}\mid +\mathrm{y}=\mathrm{Cu}$$$$\mathrm{y}=\mathrm{u}(\mathrm{C}-\ln \mid \mathrm{y}\mid )◼$$

Answered by OlafThorendsen last updated on 15/Jul/20

$$y=u\mathrm{Y}$$$$u^2{\mathrm{Y}}^2-u(u+u\mathrm{Y})(u\frac{d\mathrm{Y}}{du}+\mathrm{Y})=0$$$$u^2{\mathrm{Y}}^2-u^3(1+\mathrm{Y})\frac{d\mathrm{Y}}{du}-u^2(1+\mathrm{Y})\mathrm{Y}=0$$$$-u^3(1+\mathrm{Y})\frac{d\mathrm{Y}}{du}-u^2\mathrm{Y}=0$$$$\frac{1+\mathrm{Y}}{\mathrm{Y}}d\mathrm{Y}=-\frac{du}{u}$$$$\ln \mid \mathrm{Y}\mid +\mathrm{Y}=-\ln \mid u\mid +\mathrm{C}$$$$\ln \mid u\mid =-\ln \mid \mathrm{Y}\mid -\mathrm{Y}+\mathrm{C}$$$$\mid u\mid =\frac{\mathrm{K}{e}^{-\mathrm{Y}}}{\mid \mathrm{Y}\mid }=\frac{\mathrm{K}{e}^{-\frac{y}{u}}}{\mid \frac{y}{u}\mid }$$$$u^2=\frac{\mathrm{K}{e}^{-\frac{y}{u}}}{y}$$

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