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Question Number 129283 by math178 last updated on 14/Jan/21

Commented by math178 last updated on 14/Jan/21

$$Bernoullidifferentialequationssolution?$$$$$$

Answered by bramlexs22 last updated on 15/Jan/21

$$\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{\mathrm{y}}{\mathrm{x}}=-\frac{{\mathrm{y}}^{-3/2}}{{\mathrm{x}}^2}$$$$\mathrm{v}={\mathrm{y}}^{1-(-3/2)}={\mathrm{y}}^{5/2}$$$$\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{5}{2}{\mathrm{y}}^{3/2}\frac{\mathrm{dy}}{\mathrm{dx}}\mathrm{or}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2}{5}{\mathrm{y}}^{-3/2}\frac{\mathrm{dv}}{\mathrm{dx}}$$$$(\ast )\frac{2}{5}{\mathrm{y}}^{-3/2}\frac{\mathrm{dv}}{\mathrm{dx}}+\frac{\mathrm{y}}{\mathrm{x}}=-\frac{{\mathrm{y}}^{-3/2}}{{\mathrm{x}}^2}$$$$\frac{\mathrm{dv}}{\mathrm{dx}}+\frac{5}{2}\frac{\mathrm{v}}{\mathrm{x}}=-\frac{5}{{2\mathrm{x}}^2}$$$$\mathrm{integrating}\mathrm{factor}\gamma ={\mathrm{e}}^{\int \frac{\mathrm{dx}}{\mathrm{x}}}=\mathrm{x}$$$$\mathrm{v}=\frac{\int -\frac{5}{2\mathrm{x}}+\mathrm{C}}{\mathrm{x}}$$$${\mathrm{y}}^{5/2}=\frac{-\frac{5}{2}\ln \mathrm{x}+\mathrm{C}}{\mathrm{x}}$$$$\underset{―}{{\mathbf{y}}^{5/2}=-\frac{5\mathbf{ln}\mathbf{x}}{2\mathbf{x}}+{\mathbf{Cx}}^{-1}.}$$

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