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Question Number 99286 by bobhans last updated on 20/Jun/20

Commented by bemath last updated on 20/Jun/20

$$\mathrm{Bernoulli}\mathrm{equation}$$

Commented by bemath last updated on 20/Jun/20

$${\mathrm{y}}^{\prime }+\frac{1}{\mathrm{x}}\mathrm{y}=\frac{\cos \mathrm{x}}{{\mathrm{y}}^2}$$$${\mathrm{y}}^{\prime }+\frac{1}{\mathrm{x}}\mathrm{y}=\left(\cos \mathrm{x}\right).{\mathrm{y}}^{-2}$$$$\mathrm{let}\mathrm{u}={\mathrm{y}}^{1-(-2)}={\mathrm{y}}^3$$$$\frac{\mathrm{du}}{\mathrm{dx}}={3\mathrm{y}}^2\frac{\mathrm{dy}}{\mathrm{dx}},\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{3}{\mathrm{y}}^{-2}\frac{\mathrm{du}}{\mathrm{dx}}$$$$\iff \frac{1}{3}{\mathrm{y}}^{-2}\frac{\mathrm{du}}{\mathrm{dx}}+\frac{1}{\mathrm{x}}\mathrm{y}=\left(\cos \mathrm{x}\right).{\mathrm{y}}^{-2}$$$$\frac{\mathrm{du}}{\mathrm{dx}}+\frac{3}{\mathrm{x}}\mathrm{u}=\cos \mathrm{x}$$$$\mathrm{IF}\mathrm{v}\left(\mathrm{x}\right)={\mathrm{e}}^{\int \frac{3}{\mathrm{x}}\mathrm{dx}}={\mathrm{e}}^{3\ln \left(\mathrm{x}\right)}={\mathrm{x}}^3$$$$\iff \mathrm{u}\left(\mathrm{x}\right)=\frac{\int {\mathrm{x}}^3\cos \mathrm{x}\mathrm{dx}+\mathrm{C}}{{\mathrm{x}}^3}$$$${\mathrm{x}}^3{\mathrm{y}}^3={\mathrm{x}}^3\sin \mathrm{x}+{3\mathrm{x}}^2\cos \mathrm{x}-6\mathrm{xsin}\mathrm{x}-6\cos \mathrm{x}+\mathrm{C}$$$${\mathrm{y}}^3=\sin \mathrm{x}+\frac{3\cos \mathrm{x}}{\mathrm{x}}-\frac{6\sin \mathrm{x}}{{\mathrm{x}}^2}-\frac{6\cos \mathrm{x}}{{\mathrm{x}}^3}+{\mathrm{Cx}}^{-3}$$

Commented by bobhans last updated on 20/Jun/20

$$\mathrm{great}$$

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