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Question Number 128689 by bemath last updated on 09/Jan/21

$$({\mathrm{x}}^3+{\mathrm{y}}^3)\mathrm{dx}={3\mathrm{xy}}^2\mathrm{dy}$$

Answered by liberty last updated on 09/Jan/21

$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{{\mathrm{x}}^3+{\mathrm{y}}^3}{{3\mathrm{xy}}^2}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{3}{\mathrm{x}}^2{\mathrm{y}}^{-2}+\frac{\mathrm{y}}{3\mathrm{x}}$$$$\mathrm{let}\mathrm{v}={\mathrm{y}}^3\Rightarrow \frac{\mathrm{dv}}{\mathrm{dx}}={3\mathrm{y}}^2\frac{\mathrm{dy}}{\mathrm{dx}}\mathrm{or}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{3}{\mathrm{y}}^{-2}\frac{\mathrm{dv}}{\mathrm{dx}}$$$$\mathrm{then}\frac{1}{3}{\mathrm{y}}^{-2}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{1}{3}{\mathrm{x}}^2{\mathrm{y}}^{-2}+\frac{\mathrm{y}}{3\mathrm{x}}$$$$\iff \frac{\mathrm{dv}}{\mathrm{dx}}={\mathrm{x}}^2+\frac{{\mathrm{y}}^3}{\mathrm{x}};\frac{\mathrm{dv}}{\mathrm{dx}}-\frac{\mathrm{v}}{\mathrm{x}}={\mathrm{x}}^2$$$$\mathrm{put}\mathrm{integrating}\mathrm{factor}\mu ={\mathrm{e}}^{\int -\frac{\mathrm{dx}}{\mathrm{x}}}=\frac{1}{\mathrm{x}}$$$$\mathrm{we}\mathrm{get}\mathrm{v}=\frac{\int {\mathrm{x}}^2\left(\frac{1}{\mathrm{x}}\right)\mathrm{dx}+\mathrm{C}}{\left(\frac{1}{\mathrm{x}}\right)}$$$$\mathrm{v}=\mathrm{x}(\frac{1}{2}{\mathrm{x}}^2+\mathrm{C})\Rightarrow \mathrm{y}=\sqrt[3]{\frac{1}{2}{\mathrm{x}}^3+\mathrm{Cx}}$$$$\iff \mathrm{y}=\frac{1}{2}\sqrt[3]{{4\mathrm{x}}^3+\lambda \mathrm{x}}.$$

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