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Question Number 101846 by bemath last updated on 05/Jul/20

Answered by bramlex last updated on 05/Jul/20

$$\iff \frac{dy}{dx}-2x^{-1}y=x^2\cos x$$$$integratingfactoru=e^{\int -2x^{-1}dx}$$$$u=e{}^{-2\ln \left(x\right)}=x^{-2}$$$$\iff \mathrm{y}=\frac{\int x^{-2}.x^2\cos xdx}{x^{-2}}$$$$\iff \mathrm{y}=x^2\{\sin x+C\}$$$$\mathrm{y}=x^2\sin x+{Cx}^2$$$$(B\mathrm{ram}-◻)$$

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

$$\frac{1}{\mathrm{x}}{\mathrm{y}}^{{}^{\prime }}-\frac{2}{{\mathrm{x}}^2}\mathrm{y}=\mathrm{xcosx}\Rightarrow {\mathrm{xy}}^{{}^{\prime }}-2\mathrm{y}={\mathrm{x}}^3\mathrm{cosx}(\mathrm{x}>0)$$$$\mathrm{he}\to {\mathrm{xy}}^{{}^{\prime }}-2\mathrm{y}=0\Rightarrow {\mathrm{xy}}^{{}^{\prime }}=2\mathrm{y}\Rightarrow \frac{{\mathrm{y}}^{{}^{\prime }}}{\mathrm{y}}=\frac{2}{\mathrm{x}}\Rightarrow \ln \mid \mathrm{y}\mid =2\mathrm{lnx}+\mathrm{c}\Rightarrow$$$$\mathrm{y}=\mathrm{k}{\mathrm{x}}^2\mathrm{lsgrange}\mathrm{method}\to {\mathrm{y}}^{{}^{\prime }}={\mathrm{k}}^{{}^{\prime }}{\mathrm{x}}^2+2\mathrm{kx}$$$$\mathrm{e}\Rightarrow {\mathrm{k}}^{{}^{\prime }}{\mathrm{x}}^3+{2\mathrm{kx}}^2-{2\mathrm{kx}}^2={\mathrm{x}}^3\mathrm{cosx}\Rightarrow {\mathrm{k}}^{{}^{\prime }}=\mathrm{cosx}\Rightarrow \mathrm{k}=\mathrm{sinx}+\lambda \Rightarrow$$$$\mathrm{y}\left(\mathrm{x}\right)=(\mathrm{sinx}+\lambda ){\mathrm{x}}^2=\lambda {\mathrm{x}}^2+{\mathrm{x}}^2\mathrm{sinx}$$

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