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Question Number 85498 by Roland Mbunwe last updated on 22/Mar/20

$$(1-x^2)\frac{dy}{dx}-2xy=0$$

Commented by niroj last updated on 22/Mar/20

$$(1-{\mathrm{x}}^2)\frac{\mathrm{dy}}{\mathrm{dx}}-2\mathrm{xy}=0$$$$(1-{\mathrm{x}}^2)\frac{\mathrm{dy}}{\mathrm{dx}}=2\mathrm{xy}$$$$\frac{1}{\mathrm{y}}\mathrm{dy}=\frac{2\mathrm{x}}{1-{\mathrm{x}}^2}\mathrm{dx}$$$$\mathrm{On}\mathrm{integrating}\mathrm{both}\mathrm{side},$$$$\int \frac{1}{\mathrm{y}}\mathrm{dy}=2\int \frac{\mathrm{xdx}}{1-{\mathrm{x}}^2}$$$$\mathrm{put},1-{\mathrm{x}}^2=\mathrm{t}$$$$-2\mathrm{xdx}=\mathrm{dt}$$$$\mathrm{xdx}=-\frac{\mathrm{dt}}{2}$$$$\mathrm{logy}=-2\int \frac{\mathrm{dt}}{\mathrm{t}.2}$$$$\mathrm{logy}=-\log \mathrm{t}+\log \mathrm{c}$$$$\log \mathrm{y}=-\log (1-{\mathrm{x}}^2)+\log \mathrm{c}$$$$\log \mathrm{y}=\log \frac{\mathrm{c}}{1-{\mathrm{x}}^2}$$$$\mathrm{y}=\frac{\mathrm{c}}{1-{\mathrm{x}}^2}$$$$\mathrm{y}-{\mathrm{x}}^2\mathrm{y}=\mathrm{c}$$$$\mathrm{y}={\mathrm{x}}^2\mathrm{y}+\mathrm{C}//.$$$$$$$$$$

Answered by mind is power last updated on 22/Mar/20

$$\iff \frac{dy}{y}=\frac{2x}{1-x^2}$$$$\Rightarrow ln\mid y\mid =-ln\mid x^2-1\mid$$$$\mid y\mid =\frac{c}{\mid x^2-1\mid }$$$$\Rightarrow y=\frac{c}{\mid x^2-1\mid }$$

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