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Question Number 101848 by bramlex last updated on 05/Jul/20

$$\left(\cos \mathrm{x}\right)\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}\sin \mathrm{x}=2\mathrm{x}\cos {}^2\mathrm{x},$$$$\mathrm{y}\left(\frac{\pi }{4}\right)=\frac{-15{\pi }^2\sqrt{2}}{32}$$

Answered by john santu last updated on 05/Jul/20

$$\frac{dy}{dx}+y.\tan \left(x\right)=2x\cos x$$$$\Rightarrow IF:u\left(x\right)=e^{\int \tan xdx}=e^{-\int \frac{d\cos x}{\cos x}}$$$$u\left(x\right)=e^{-\ln \left(\cos x\right)}=\frac{1}{\cos x}$$$$\Rightarrow \frac{dy}{\cos x.dx}+\frac{y\sin x}{\cos {}^{2}x}=2x$$$$\frac{d}{dx}\left(\frac{y}{\cos x}\right)=2x\Rightarrow \frac{y}{\cos x}=x^2+C$$$$y\left(x\right)=\cos x(x^2+C)$$$$\frac{-15{\pi }^2\sqrt{2}}{32}=\frac{\sqrt{2}}{2}(\frac{{\pi }^2}{16}+C)$$$$\Rightarrow -\frac{15{\pi }^2}{16}=\frac{{\pi }^2}{16}+C$$$$C=-{\pi }^2$$$$\therefore solutiony\left(x\right)=(x^2-{\pi }^2).\cos x$$$$(JS⊛)$$

Commented by bramlex last updated on 05/Jul/20

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Answered by mathmax by abdo last updated on 05/Jul/20

$$\mathrm{cosx}{\mathrm{y}}^{{}^{\prime }}+\mathrm{sinx}\mathrm{y}={2\mathrm{xcos}}^2\mathrm{x}\mathrm{with}\mathrm{y}\left(\frac{\pi }{4}\right)=-\frac{15\pi \sqrt{2}}{32}$$$$\mathrm{he}\to \mathrm{cosx}{\mathrm{y}}^{{}^{\prime }}+\mathrm{sinx}\mathrm{y}=0\Rightarrow \mathrm{cosx}{\mathrm{y}}^{{}^{\prime }}=-\mathrm{sinx}\mathrm{y}\Rightarrow \frac{{\mathrm{y}}^{{}^{\prime }}}{\mathrm{y}}=-\frac{\mathrm{sinx}}{\mathrm{cosx}}\Rightarrow$$$$\ln \mid \mathrm{y}\mid =\ln \left(\mathrm{cosx}\right)+\mathrm{c}\Rightarrow \mathrm{y}=\mathrm{k}\mathrm{cosx}$$$$\mathrm{lagrange}\mathrm{method}\to {\mathrm{y}}^{{}^{\prime }}={\mathrm{k}}^{{}^{\prime }}\mathrm{cosx}-\mathrm{ksinx}$$$$\mathrm{e}\Rightarrow {\mathrm{k}}^{{}^{\prime }}{\cos }^2\mathrm{x}-\mathrm{kcosx}\mathrm{sinx}+\mathrm{ksinx}\mathrm{cosx}={2\mathrm{xcos}}^2\mathrm{x}\Rightarrow {\mathrm{k}}^{{}^{\prime }}=2\mathrm{x}\Rightarrow \mathrm{k}\left(\mathrm{x}\right)={\mathrm{x}}^2+\lambda \Rightarrow$$$$\mathrm{y}\left(\mathrm{x}\right)=({\mathrm{x}}^2+\lambda )\mathrm{cosx}$$$$\mathrm{y}\left(\frac{\pi }{4}\right)=-\frac{15\pi \sqrt{2}}{32}\Rightarrow (\frac{{\pi }^2}{8}+\lambda )\times \frac{\sqrt{2}}{2}=-\frac{15\pi \sqrt{2}}{32}\Rightarrow \lambda +\frac{{\pi }^2}{8}=-\frac{15\pi }{16}\Rightarrow$$$$\lambda =-\frac{{\pi }^2}{8}-\frac{15\pi }{16}=-\frac{17\pi }{16}\Rightarrow \mathrm{y}\left(\mathrm{x}\right)=({\mathrm{x}}^2-\frac{17\pi }{16})\mathrm{cosx}$$

Commented by john santu last updated on 05/Jul/20

$$sirx=\frac{\pi }{4}then{x}^2=\frac{{\pi }^2}{16}.?$$$$$$

Commented by mathmax by abdo last updated on 05/Jul/20

$$\mathrm{sorry}\mathrm{in}\mathrm{the}\mathrm{qustion}\mathrm{y}\left(\frac{\pi }{4}\right)=-\frac{15{\pi }^2\sqrt{2}}{32}\Rightarrow (\frac{{\pi }^2}{16}+\lambda )\times \frac{\sqrt{2}}{2}=-\frac{15{\pi }^2\sqrt{2}}{32}\Rightarrow$$$$\frac{{\pi }^2}{16}+\lambda =-\frac{15{\pi }^2}{16}\Rightarrow \lambda =-{\pi }^2\Rightarrow \mathrm{y}\left(\mathrm{x}\right)=({\mathrm{x}}^2-{\pi }^2)\mathrm{cosx}$$

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