The solution of differential equation $\cot y dx=xdy$ is ............ .
$$\begin{aligned} &\text { Given differential equation is }\\ &\begin{aligned} \cot y d x & =x d y \\ \Rightarrow\quad \frac{1}{x} d x & =\tan y d y \end{aligned} \end{aligned}$$
On integrating both sides, we get
$$\begin{array}{ll} \Rightarrow & \int \frac{1}{x} d x=\int \tan y d y \\ \Rightarrow & \log (x)=\log (\sec y)+\log C \end{array}$$
$\Rightarrow \quad \log \left(\frac{x}{\sec y}\right)=\log C$
$$\begin{aligned} \Rightarrow \quad& \frac{x}{\sec y} =C \\ \Rightarrow \quad & x =C \sec y \end{aligned}$$
The integrating factor of $\frac{d y}{d x}+y=\frac{1+y}{x}$ is ............. .
Given differential equation is
$$\begin{aligned} &\text { Given differential equation is }\\ &\begin{aligned} \frac{d y}{d x}+y & =\frac{1+y}{x} \\ \frac{d y}{d x}+y & =\frac{1}{x}+\frac{y}{x} \\ \Rightarrow \quad \frac{d y}{d x}+y\left(1-\frac{1}{x}\right) & =\frac{1}{x} \end{aligned} \end{aligned}$$
$$\begin{aligned} \therefore\quad \mathrm{IF} & =e^{\int\left(1-\frac{1}{x}\right) d x} \\ & =e^{x-\log x} \\ & =e^x \cdot e^{-\log x}=\frac{e^x}{x} \end{aligned}$$
Integrating factor of the differential of the form $\frac{d x}{d y}+P_1 x=Q_1$ is given by $e^{\int P_1 d y}$.
Solution of the differential equation of the type $\frac{d x}{d y}+P_1 x=Q_1$ is given by $x \cdot \mathrm{IF}=\int(\mathrm{IF}) \times Q_1 d y$.
Correct substitution for the solution of the differential equation of the type $\frac{d y}{d x}=f(x, y)$, where $f(x, y)$ is a homogeneous function of zero degree is $y=v x$.