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Question Number 96192 by 1549442205 last updated on 30/May/20

$$\mathrm{find}\mathrm{a}\mathrm{particular}\mathrm{solution}\mathrm{to}\mathrm{the}\mathrm{equation}$$$${\mathrm{y}}^{\prime }=\frac{\mathrm{y}}{\mathrm{x}}+\sin \frac{\mathrm{y}}{\mathrm{x}}\mathrm{with}\mathrm{original}\mathrm{condition}$$$$\mathrm{y}\left(1\right)=\frac{\pi }{2}$$

Commented by john santu last updated on 30/May/20

$$\mathrm{set}v=\frac{y}{x}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=v+x\frac{dv}{dx}$$$$v+x\frac{dv}{dx}=v+\sin v$$$$x\frac{dv}{dx}=\sin v\Rightarrow \int \frac{dv}{\sin v}=\int \frac{dx}{x}$$$$\int \csc vdv=\ln \left(x\right)+c$$$$\ln (\csc v-\cot v)=\ln \mathrm{C}x$$$$cscv-\cot v=\mathrm{C}x$$$$\frac{1-\cos \left(\frac{y}{x}\right)}{\sin \left(\frac{y}{x}\right)}=Cx$$$$1-\cos \left(\frac{y}{x}\right)=Cx\sin \left(\frac{y}{x}\right)$$$$$$

Commented by 1549442205 last updated on 11/Jun/20

$$\mathrm{This}\mathrm{is}\mathrm{an}\mathrm{other}\mathrm{way}$$$$\mathrm{Put}\frac{\mathrm{y}}{\mathrm{x}}=\mathrm{t}\Rightarrow \mathrm{y}=\mathrm{tx}\Rightarrow \mathrm{dy}=\mathrm{xdt}+\mathrm{tdx}.\mathrm{Hence},$$$$\mathrm{xdt}+\mathrm{tdx}=(\mathrm{t}+\mathrm{sint})\mathrm{dx}\Rightarrow \mathrm{xdt}=\mathrm{sintdx}$$$$\Rightarrow \frac{\mathrm{dt}}{\mathrm{sint}}=\frac{\mathrm{dx}}{\mathrm{x}}.\mathrm{Integrate}\mathrm{two}\mathrm{sides}\mathrm{we}\mathrm{get}$$$$\ln \mid \tan \left(\frac{\mathrm{t}}{2}\right)\mid =\ln \mid \mathrm{x}\mid +\mathrm{lnC}.\mathrm{From}\mathrm{that}\mid$$$$\frac{\mathrm{t}}{2}=\arctan \left(\mathrm{Cx}\right)\Rightarrow \mathrm{y}=2\mathrm{x}.\arctan \left(\mathrm{Cx}\right)$$$$\mathrm{Using}\mathrm{the}\mathrm{original}\mathrm{condition}\mathrm{we}\mathrm{get}$$$$\frac{\pi }{2}=2\mathrm{arctanC}\Rightarrow \mathrm{C}=1.\mathrm{Thus},\mathrm{the}$$$$\mathrm{particular}\mathrm{solution}\mathrm{has}\mathrm{form}:$$$$\mathbf{y}=2\mathbf{x}.\mathbf{arctanx}$$

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