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Question Number 59872 by necx1 last updated on 15/May/19

$$solvetheode$$$$(1+siny)dx=\{2y\cos y-x(secy+tany)\}dy$$

Answered by ajfour last updated on 15/May/19

$$\frac{\mathrm{dx}}{\mathrm{dy}}+\mathrm{x}(\sec \mathrm{y}+\tan \mathrm{y})=2\mathrm{ycos}\mathrm{y}$$$${\mathrm{e}}^{\int \frac{1+\sin \mathrm{y}}{\cos \mathrm{y}}\mathrm{dy}}={\mathrm{e}}^{\int \tan \frac{\mathrm{y}}{2}\mathrm{dy}}={\mathrm{e}}^{2\ln \sec \frac{\mathrm{y}}{2}}$$$$=\sec {}^2\frac{\mathrm{y}}{2}$$$$\mathrm{xsec}{}^2\frac{\mathrm{y}}{2}=\int \frac{4\mathrm{ycos}\mathrm{y}}{(1+\cos \mathrm{y})}\mathrm{dy}$$$$\mathrm{xsec}{}^2\frac{\mathrm{y}}{2}=4\int \mathrm{ydy}-4\int \frac{\mathrm{ydy}}{1+\cos \mathrm{y}}$$$$\mathrm{xsec}{}^2\frac{\mathrm{y}}{2}=4\int \mathrm{ydy}-2\int \mathrm{ysec}{}^2\frac{\mathrm{y}}{2}\mathrm{dy}$$$$\mathrm{xsec}{}^2\frac{\mathrm{y}}{2}={2\mathrm{y}}^2-2[2\mathrm{ytan}\frac{\mathrm{y}}{2}-\int 2\tan \frac{\mathrm{y}}{2}\mathrm{dy}]$$$$x\sec {}^2\frac{y}{2}=2y^2-4y\tan \frac{y}{2}+8\ln \left(\sec \frac{y}{2}\right)+c.$$$$$$

Commented by necx1 last updated on 18/May/19

$$yeah.....Sograteful$$

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