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Question Number 102653 by Sontsaronald last updated on 10/Jul/20

Answered by bemath last updated on 10/Jul/20

$$y^{\prime }-y\tan x=\sin x$$$$IFu\left(x\right)=e^{-\int tanxdx}=$$$$e^{\ln \left(\cos x\right)}=\cos x$$$$y\left(x\right)=\frac{\int \cos x\sin xdx+C}{\cos x}$$$$y\left(x\right)=\sec x(-\frac{1}{4}\cos 2x+C)$$

Answered by Ar Brandon last updated on 10/Jul/20

$${\mathrm{y}}^{\prime }=\mathrm{ytan}\left(\mathrm{x}\right)+\sin \left(\mathrm{x}\right)$$$$$$$$\Rightarrow \{{\mathrm{y}}^{\prime }-\mathrm{ytan}\left(\mathrm{x}\right)=\sin \left(\mathrm{x}\right)\}.\cos \left(\mathrm{x}\right)$$$$$$$$\Rightarrow {\mathrm{y}}^{\prime }\cos \left(\mathrm{x}\right)-\mathrm{ysin}\left(\mathrm{x}\right)=\sin \left(\mathrm{x}\right)\cos \left(\mathrm{x}\right)$$$$$$$$\Rightarrow \frac{\mathrm{d}\left(\mathrm{ycos}\left(\mathrm{x}\right)\right)}{\mathrm{dx}}=\sin \left(\mathrm{x}\right)\cos \left(\mathrm{x}\right)$$$$\Rightarrow \mathrm{ycos}\left(\mathrm{x}\right)=\int \sin \left(\mathrm{x}\right)\cos \left(\mathrm{x}\right)\mathrm{dx}=\frac{{\sin }^2\left(\mathrm{x}\right)}{2}+\mathcal{C}$$$$\Rightarrow \mathrm{y}=\frac{{\sin }^2\left(\mathrm{x}\right)+\mathrm{K}}{2\cos \left(\mathrm{x}\right)},\mathrm{K}=2\mathcal{C}$$

Commented by Sontsaronald last updated on 10/Jul/20

$$mercibeaucoup$$

Commented by Sontsaronald last updated on 10/Jul/20

$$ilvousplaitquelleestdonclabonnereponse?$$

Commented by prakash jain last updated on 10/Jul/20

$${\sin }^{2}x=\frac{1-\cos 2x}{2}$$$$\frac{{\sin }^{2}x+K}{2\cos x}=\frac{1}{\cos x}\left(\frac{1-\cos 2x+K}{2\times 2}\right)$$$$=\sec x(-\frac{\cos 2x}{4}+\mathrm{constants})$$$$\mathrm{both}\mathrm{are}\mathrm{equal}$$

Commented by Ar Brandon last updated on 10/Jul/20

Je vous en prie �� Les deux réponses sont les mêmes, comme vient de le prouver ce monsieur.��

Commented by Ar Brandon last updated on 10/Jul/20

����

Commented by Sontsaronald last updated on 10/Jul/20

$$mercibeaucoupohhh$$

Commented by Ar Brandon last updated on 10/Jul/20

Cette expression ! �� Ça me rappelle de quelqu'un.

Answered by mathmax by abdo last updated on 11/Jul/20

$${\mathrm{y}}^{{}^{\prime }}-\mathrm{ytanx}=\mathrm{sinx}$$$$\mathrm{he}\to {\mathrm{y}}^{{}^{\prime }}-\mathrm{ytanx}=0\Rightarrow {\mathrm{y}}^{{}^{\prime }}=\mathrm{ytanx}\Rightarrow \frac{{\mathrm{y}}^{{}^{\prime }}}{\mathrm{y}}=\mathrm{tanx}\Rightarrow \ln \mid \mathrm{y}\mid =\int \frac{\mathrm{sinx}}{\mathrm{cosx}}\mathrm{dx}=-\ln \mid \mathrm{cosx}\mid +\mathrm{c}$$$$\Rightarrow \mathrm{y}=\frac{\mathrm{k}}{\mid \mathrm{cosx}\mid }\mathrm{let}\mathrm{solve}\mathrm{on}\mathrm{d}=\{\mathrm{x}/\mathrm{cosx}>0\}\Rightarrow \mathrm{y}=\frac{\mathrm{k}}{\mathrm{cosx}}$$$$\mathrm{lagrange}\mathrm{method}\Rightarrow {\mathrm{y}}^{{}^{\prime }}=\frac{{\mathrm{k}}^{{}^{\prime }}}{\mathrm{cosx}}+\mathrm{k}\left(\frac{\mathrm{sinx}}{{\cos }^2\mathrm{x}}\right)$$$$\mathrm{e}\Rightarrow \frac{{\mathrm{k}}^{{}^{\prime }}}{\mathrm{cosx}}+\frac{\mathrm{ksinx}}{{\cos }^2\mathrm{x}}-\frac{\mathrm{k}}{\mathrm{cosx}}\times \frac{\mathrm{sinx}}{\mathrm{cosx}}=\mathrm{sinx}\Rightarrow \frac{{\mathrm{k}}^{{}^{\prime }}}{\mathrm{cosx}}=\mathrm{sinx}\Rightarrow {\mathrm{k}}^{{}^{\prime }}=\frac{1}{2}\sin \left(2\mathrm{x}\right)\Rightarrow$$$$\mathrm{k}\left(\mathrm{x}\right)=\frac{1}{2}\int \sin \left(2\mathrm{x}\right)\mathrm{dx}=-\frac{1}{4}\cos \left(2\mathrm{x}\right)+\lambda \Rightarrow$$$$\mathrm{y}\left(\mathrm{x}\right)=\frac{1}{\mathrm{cosx}}\{-\frac{1}{4}\cos \left(2\mathrm{x}\right)+\lambda \}$$

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