Maths Solve the Following Question.(2 Marks)

$\begin{aligned} & y-x \frac{d y}{d x}=0 \\ & \therefore y=x \frac{d y}{d x} \\ & \therefore \frac{d x}{x}=\frac{d y}{y}\end{aligned}$
Integrating on both sides, we get
$\int \frac{d x}{x}=\int \frac{d y}{y}$
∴ log |x| = log |y| + log |c|
∴ log |x| = log |cy|
∴ x = cy

$\begin{aligned} & \frac{d y}{d x}=x^2 y+y=\left(x^2+1\right) y \\ & \therefore \frac{1}{y} d y=\left(x^2+1\right) d x\end{aligned}$
Integrating on both sides, we get
$\begin{aligned} & \int \frac{1}{y} d y=\int\left(x^2+1\right) d x \\ & \therefore \log |y|=\frac{x^3}{3}+x+c\end{aligned}$

$y=x^2+4 x+1$
Differentiating w.r.t 'x', we get
$\frac{d y}{d x}=2 x+4$
$\left.\frac{d y}{d x}\right|_{x \rightarrow-1}=2(-1)+4=2$
Hence, slope of tangent at (-1, -2) is 2.
So equation of tangent line is
y -(-2)= 2( x-(- 1))
2x - y = 0

$y=c_1 e^{2 x}+c_2 e^{-2 x}$
differentiate w.r.t. x.
$\frac{d y}{d x}=2 c_1 e^{2 x}-2 c_2 e^{-2 x}$
Again diff. w.r.t. x.
$\begin{aligned} & \frac{d^2 y}{d x^2}=4 c_1 e^{2 x}+4 c_2 e^{-2 x} \\ & =4\left(c_1 e^{2 x}+c_2 e^{-2 x}\right) \\ & =4 y \\ & \therefore \frac{d^2 y}{d x^2}-4 y=0\end{aligned}$

$y=A e^{5 x}+B e^{-5 x}$
Differentitating w.r.t. x
$\begin{aligned} & \frac{d y}{d x}=A \cdot e^{5 x} \cdot(5)+B e^{-5 x}(-5) \\ & \therefore \frac{d y}{d x}=5 A \cdot e^{5 x}-5 B e^{-5 x}\end{aligned}$
Again differentitating w.r.t. x
$\begin{aligned} & \frac{d^2 y}{d x^2}=5 A e^{5 x} \cdot(5)-5 B e^{-5 x} \cdot(-5) \\ & \frac{d^2 y}{d x^2}=25 A e^{5 x}+25 B e^{-5 x} \\ & \frac{d^2 y}{d x^2}=25\left(A e^{5 x}+B e^{-5 x}\right) \\ & \frac{d^2 y}{d x^2}=25 y \\ & \frac{d^2 y}{d x^2}-25 y=0 \text { is the required differential equation }\end{aligned}$

The given differential equation is :
$\begin{aligned} & \sec ^2 x \tan y d x+\sec ^2 y \tan x d y=0 \\ & \Rightarrow \frac{\sec ^2 x \tan y d x+\sec ^2 y \tan x d y}{\tan x \tan y}=0 \\ & \Rightarrow \frac{\sec ^2 x}{\tan x} d x+\frac{\sec ^2 y}{\tan y} d y=0 \\ & \Rightarrow \frac{\sec ^2 x}{\tan x} d x=-\frac{\sec ^2 y}{\tan y} d y\end{aligned}$
Integrating both sides of this equation , we get : 
$\int \frac{\sec ^2 x}{\tan x} d x=-\int \frac{\sec ^2 y}{\tan y} d y$...(1)
Let tanx = t .
$\begin{aligned} & \therefore \frac{d}{d x}(\tan x)=\frac{d t}{d x} \\ & \Rightarrow \sec ^2 x=\frac{d t}{d x} \\ & \Rightarrow \sec ^2 x d x=d t \\ & \text { Now, } \int \frac{\sec ^2 x}{\tan x} d x=\int \frac{1}{t} d t .\end{aligned}$
= log t
= log (tan x)
Similarly, $\int \frac{\sec ^2 y}{\tan y} d y=\log (\tan y)$.
log(tan x) = - log(tan y) + log C
$\begin{aligned} & \Rightarrow \log (\tan x)+\log (\tan y)=\log C \\ & \Rightarrow \log (\tan x \tan y)=\log C \quad[\because \log a+\log b=\log (a b)] \\ & \Rightarrow \tan x \tan y=C\end{aligned}$