The general solution of $\sin y \cdot \frac{d y}{d x}=\cos y(1-x \cos y)$ is

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Given differential equation is $\begin{array}{rlrl} \Rightarrow \quad & & \sin y \frac{d y}{d x} & =\cos y(1-x \cos y) \ \Rightarrow & \quad \sin y \frac{d y}{d x} & =\cos y-x \cos ^2 y \ \Rightarrow \quad & \quad \frac{\sin y}{\cos ^2 y} \frac{d y}{d x} & =\frac{1}{\cos y}-x \ \Rightarrow \quad & \sec y \tan y \frac{d y}{d x} & =\sec y-x \end{array}$ Let $\quad \sec y=t$ $\begin{aligned} & \Rightarrow \sec y \tan y \frac{d y}{d x}=\frac{d t}{d x} \ & \therefore \quad \frac{d t}{d x}=t-x \ & \Rightarrow \quad \frac{d t}{d x}+(-t)=-x \ & \therefore \quad \mathrm{IF}=e^{\int-d x}=e^{-x} \end{aligned}$ Now, required solution $\begin{aligned} \Rightarrow \quad t(I F) & =\int(-x)(I F) d x+c \ \sec y\left(e^{-x}\right) & =\int(-x) e^{-x} d x+c \ & =\left(x e^{-x}\right)-\int e^{-x} d x+c \ & =x e^{-x}+e^{-x}+c \ \Rightarrow \quad & \sec y=x+1+c e^x \end{aligned}$

The general solution of \(\sin y \cdot \frac{d y}{d x}=\cos y(1-x \cos y)\) is

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