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Question Number 40397 by rahul 19 last updated on 21/Jul/18

$$\mathrm{Solve}:$$$$\frac{\mathrm{dy}}{dx}=\frac{\sin y+x}{\sin 2y-x\cos y}.$$

Answered by ajfour last updated on 21/Jul/18

$$\cos y\left(\frac{dy}{dx}\right)=\frac{x+\sin y}{2\sin y-x}$$$$let\sin y=t$$$$\Rightarrow \left(\cos y\right)\frac{dy}{dx}=\frac{dt}{dx}$$$$So\frac{dt}{dx}=\frac{x+t}{2t-x}$$$$\Rightarrow 2tdt-xdt=xdx+tdx$$$$2tdt-xdx=d\left(tx\right)$$$$\Rightarrow t^2-\frac{x^2}{2}=tx+\frac{c}{2}$$$$2\sin {}^{2}y-x^2=2x\sin y+c.$$

Commented by rahul 19 last updated on 21/Jul/18

thank you sir ��

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