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Question Number 116425 by bemath last updated on 04/Oct/20

$${\mathrm{y}}^{\prime }-\mathrm{y}=-{2\mathrm{xy}}^3$$

Commented by bemath last updated on 04/Oct/20

$$\mathrm{thank}\mathrm{you}\mathrm{mr}\mathrm{Bob}\mathrm{and}\mathrm{mr}\mathrm{Olaf}$$

Answered by bobhans last updated on 04/Oct/20

$${\mathrm{y}}^{\prime }-\mathrm{y}=-{2\mathrm{xy}}^3\leftarrow \mathrm{Bernoulli}\mathrm{Diff}\mathrm{eq}$$$$\mathrm{let}\mathrm{v}={\mathrm{y}}^{-2}\Rightarrow \frac{\mathrm{dv}}{\mathrm{dx}}=-{2\mathrm{y}}^{-3}\frac{\mathrm{dy}}{\mathrm{dx}}$$$$\mathrm{and}\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{1}{2}{\mathrm{y}}^3\frac{\mathrm{dv}}{\mathrm{dx}}$$$$\mathrm{let}\mathrm{we}\mathrm{consider}$$$$\Rightarrow -\frac{1}{2}{\mathrm{y}}^3\frac{\mathrm{dv}}{\mathrm{dx}}-\mathrm{y}=-{2\mathrm{xy}}^3$$$$\Rightarrow -\frac{1}{2}\frac{\mathrm{dv}}{\mathrm{dx}}-\mathrm{v}=-2\mathrm{x}$$$$\Rightarrow \frac{\mathrm{dv}}{\mathrm{dx}}+2\mathrm{v}=4\mathrm{x};\mathrm{Integrating}\mathrm{factor}$$$$\mathrm{u}={\mathrm{e}}^{\int 2\mathrm{dx}}={\mathrm{e}}^{2\mathrm{x}}$$$$\Rightarrow \mathrm{v}=\frac{\int {\mathrm{e}}^{2\mathrm{x}}.4\mathrm{x}\mathrm{dx}+\mathrm{C}}{{\mathrm{e}}^{2\mathrm{x}}}$$$$\Rightarrow \frac{1}{{\mathrm{y}}^2}=\frac{2\mathrm{x}.{\mathrm{e}}^{2\mathrm{x}}-{\mathrm{e}}^{2\mathrm{x}}+\mathrm{C}}{{\mathrm{e}}^{2\mathrm{x}}}=2\mathrm{x}-1+\mathrm{C}.{\mathrm{e}}^{-2\mathrm{x}}$$$$\Rightarrow \mathrm{y}=\pm \frac{1}{\sqrt{\mathrm{C}.{\mathrm{e}}^{-2\mathrm{x}}+2\mathrm{x}-1}}$$$$$$

Answered by Olaf last updated on 04/Oct/20

$$\mathrm{I}\mathrm{suppose}\mathrm{that}y\mathrm{is}\mathrm{a}\mathrm{positive}\mathrm{function}$$$$\mathrm{and}y=\frac{1}{\sqrt{u}}$$$$-\frac{1}{2}.\frac{u^{\prime }}{u^{3/2}}-\frac{1}{u^{1/2}}=-2x\frac{1}{u^{3/2}}$$$$u^{\prime }+2u=4x$$$$u_0=2x-1\mathrm{is}\mathrm{a}\mathrm{particular}\mathrm{solution}.$$$$\mathrm{Homogen}\mathrm{equation}:$$$$u_{\mathrm{H}}^{\prime }+2u_{\mathrm{H}}=0$$$$\frac{u_{\mathrm{H}}^{\prime }}{u_{\mathrm{H}}}=-2$$$$\ln u_{\mathrm{H}}=-2x+\mathrm{C}$$$$u_{\mathrm{H}}=\mathrm{K}{e}^{-2x}$$$$u=u_{\mathrm{H}}+u_0=\mathrm{K}{e}^{-2x}+2x-1$$$$y=\frac{1}{\sqrt{u}}=\frac{1}{\sqrt{\mathrm{K}{e}^{-2x}+2x-1}}$$$$\mathrm{Symetrically},\mathrm{if}y\mathrm{is}\mathrm{a}\mathrm{negative}\mathrm{function}$$$$y=-\frac{1}{\sqrt{\mathrm{K}{e}^{-2x}+2x-1}}\mathrm{is}\mathrm{a}\mathrm{solution}\mathrm{too}.$$$$$$

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