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Question Number 85513 by niroj last updated on 22/Mar/20

$$\mathbf{Solve}\mathbf{the}\mathbf{following}\mathbf{equation}:$$$${\left(\frac{\mathbf{dy}}{\mathbf{dx}}\right)}^2+2\mathbf{y}\mathbf{cot}\mathbf{x}\frac{\mathbf{dy}}{\mathbf{dx}}={\mathbf{y}}^2$$

Answered by mr W last updated on 22/Mar/20

$$\frac{dy}{dx}=-y\cot x\pm \sqrt{y^2{\cot }^{2}x+y^2}$$$$\frac{dy}{dx}=-\frac{(\cos x\pm 1)y}{\sin x}$$$$\frac{dy}{y}=-\frac{(\cos x\pm 1)dx}{\sin x}$$$$\int \frac{dy}{y}=-\int \frac{(\cos x\pm 1)dx}{\sin x}$$$$\int \frac{dy}{y}=\int \frac{(\cos x\pm 1)d\left(\cos x\right)}{1-{\cos }^{2}x}$$$$\ln y=\int \frac{(u\pm 1)du}{1-u^2}$$$$$$$$\ln y=\int \frac{(u+1)du}{1-u^2}$$$$\ln y=\int \frac{du}{1-u}$$$$\ln y=-\ln (1-u)+C$$$$\ln y(1-\cos x)=C$$$$\Rightarrow y=\frac{C}{1-\cos x}$$$$$$$$\ln y=\int \frac{(u-1)du}{1-u^2}$$$$\ln y=-\int \frac{du}{1+u}$$$$\ln y=-\ln (1+u)+C$$$$\ln y(1+\cos x)=C$$$$\Rightarrow y=\frac{C}{1+\cos x}$$

Commented by mr W last updated on 22/Mar/20

$$check:$$$$y=\frac{C}{1+\cos x}$$$$\frac{dy}{dx}=\frac{C\sin x}{{(1+\cos x)}^2}$$$${\left(\frac{dy}{dx}\right)}^2=\frac{C^2{\sin }^{2}x}{{(1+\cos x)}^4}=\frac{C^2(1-\cos x)}{{(1+\cos x)}^3}$$$${\left(\frac{dy}{dx}\right)}^2+2y\cot x\left(\frac{dy}{dx}\right)=\frac{C^2(1-\cos x)}{{(1+\cos x)}^3}+2\left(\frac{C}{1+\cos x}\right)\frac{\cos x}{\sin x}\left(\frac{C\sin x}{{(1+\cos x)}^2}\right)$$$$=\frac{C^2(1-\cos x)}{{(1+\cos x)}^3}+\frac{2C^2\cos x}{{(1+\cos x)}^3}$$$$=\frac{C^2}{{(1+\cos x)}^2}$$$$=y^2$$

Commented by niroj last updated on 22/Mar/20

$$\mathrm{thank}\mathrm{you}\mathrm{mr}.\mathrm{w}nicesolution.$$

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