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Question Number 116614 by bemath last updated on 05/Oct/20

$$({\mathrm{x}}^2\cos {}^2\left(\frac{\mathrm{x}}{\mathrm{y}}\right)-\mathrm{y})\mathrm{dx}+\mathrm{x}\mathrm{dy}=0$$$$\mathrm{where}\mathrm{y}\left(1\right)=\frac{\pi }{4}$$

Commented by TANMAY PANACEA last updated on 05/Oct/20

$$ithinkitshouldbe{cos}^2\left(\frac{y}{x}\right)$$

Commented by bobhans last updated on 05/Oct/20

$$\mathrm{let}\frac{\mathrm{x}}{\mathrm{y}}=\mathrm{v}\Rightarrow \mathrm{y}={\mathrm{xv}}^{-1}$$$$\frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{v}}^{-1}-{\mathrm{xv}}^{-2}\frac{\mathrm{dv}}{\mathrm{dx}}$$$$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}-{\mathrm{x}}^2\cos {}^2\left(\frac{\mathrm{x}}{\mathrm{y}}\right)}{\mathrm{x}}$$$$\Rightarrow \frac{1}{\mathrm{v}}-\frac{\mathrm{x}}{{\mathrm{v}}^2}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\frac{\mathrm{x}}{\mathrm{v}}-{\mathrm{x}}^2\cos {}^2\left(\mathrm{v}\right)}{\mathrm{x}}$$$$\Rightarrow \frac{1}{\mathrm{v}}-\frac{\mathrm{x}\mathrm{dv}}{{\mathrm{v}}^2\mathrm{dx}}=\frac{1}{\mathrm{v}}-\mathrm{x}\cos {}^2\left(\mathrm{v}\right)$$$$\Rightarrow \frac{\mathrm{x}\mathrm{dv}}{{\mathrm{v}}^2\mathrm{dx}}=\mathrm{x}\cos {}^2\left(\mathrm{v}\right)$$$$\Rightarrow \frac{\mathrm{dv}}{\cos {}^2\left(\mathrm{v}\right)}=\mathrm{dx};\int \sec {}^2\left(\mathrm{v}\right)\mathrm{dv}=\int \mathrm{dx}$$$$\Rightarrow \tan \left(\mathrm{v}\right)=\mathrm{x}+\mathrm{c}$$$$\Rightarrow \tan \left(\frac{\mathrm{x}}{\mathrm{y}}\right)=\mathrm{x}+\mathrm{c},\mathrm{from}\mathrm{given}\mathrm{condition}$$$$\mathrm{y}\left(1\right)=\frac{\pi }{4}\Rightarrow \tan \left(\frac{4}{\pi }\right)=1+\mathrm{c}$$$$\Rightarrow 0.022-1=\mathrm{c},\mathrm{c}=-0.978$$$$\mathrm{therefore}\mathrm{solution}\mathrm{is}\frac{\mathrm{x}}{\mathrm{y}}={\tan }^{-1}(\mathrm{x}-0.978)$$$$\mathrm{or}\mathrm{y}=\frac{\mathrm{x}}{{\tan }^{-1}(\mathrm{x}-0.978)}$$

Commented by bemath last updated on 05/Oct/20

$$\mathrm{sir}\mathrm{Tanmay}.\mathrm{in}\mathrm{my}\mathrm{book}\mathrm{the}\mathrm{question}$$$$\mathrm{correct}\mathrm{sir}$$

Answered by TANMAY PANACEA last updated on 05/Oct/20

$$rechecktheproblempls$$

Answered by TANMAY PANACEA last updated on 05/Oct/20

$$assume{\mathit{c}\mathit{o}\mathit{s}}^2\left(\frac{\mathit{y}}{\mathit{x}}\right)\mathit{g}\mathit{i}\mathit{v}\mathit{e}\mathit{n}$$$$\mathit{x}\mathit{d}\mathit{y}-\mathit{y}\mathit{d}\mathit{x}+{\mathit{x}}^2{\mathit{c}\mathit{o}\mathit{s}}^2\left(\frac{\mathit{y}}{\mathit{x}}\right)\mathit{d}\mathit{x}=0$$$$\frac{\mathit{x}\mathit{d}\mathit{y}-\mathit{y}\mathit{d}\mathit{x}}{{\mathit{x}}^2}\times {\mathit{s}\mathit{e}\mathit{c}}^2\left(\frac{\mathit{y}}{\mathit{x}}\right)+\mathit{d}\mathit{x}=\mathit{d}\mathit{c}$$$${\mathit{s}\mathit{e}\mathit{c}}^2\left(\frac{\mathit{y}}{\mathit{x}}\right)\mathit{d}\left(\frac{\mathit{y}}{\mathit{x}}\right)+\mathit{d}\mathit{x}=\mathit{d}\mathit{c}$$$$\mathit{i}\mathit{n}\mathit{t}\mathit{r}\mathit{e}\mathit{g}\mathit{a}\mathit{t}\mathit{i}\mathit{n}\mathit{g}$$$$\mathit{t}\mathit{a}\mathit{n}\left(\frac{\mathit{y}}{\mathit{x}}\right)+\mathit{x}=\mathit{c}$$

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