Maths Solve the Following Question.(3 Marks)

$x+y \frac{d y}{d x}=\sec \left(x^2+y^2\right)$
Put $x^2+y^2=v \Rightarrow 2 x+2 y \frac{d y}{d x}=\frac{d v}{d x}$
$x+y \frac{d y}{d x}=\frac{1}{2} \frac{d v}{d x}$
$\therefore$ Given equation becomes,
$\frac{1}{2} \frac{d v}{d x}=\sec v$
$\therefore \cos v d v=2 d x$
$\therefore$ Integrating both sides,
$\int \cos v d v=2 \int d x$
$\therefore \sin v=2 x+c$
Substituting the value of $v$,
$\sin \left(x^2+y^2\right)=2 x+c$

Let P be the population of the country at time t.
$\begin{aligned} & \text { Given } \frac{ dP }{d t} \propto P \\ & \therefore \frac{ dP }{d t}=k P \text { (where } k \text { is a constant) } \\ & \therefore \frac{1}{P} d P=k d t\end{aligned}$
Integrating both the side w.r.t x
$\begin{aligned} & \int \frac{1}{P} d p=k \int 1 d t+c \\ & \log P=k t+c \\ & P=e^{k t+c}=e^{k t} \cdot e^c \\ & \text { Let } e^c=\alpha \\ & \therefore P=\alpha \cdot e^{k t}\end{aligned}$
Let initial population at t = 0
$\begin{aligned} & \therefore N=\alpha \cdot e^0 \quad \therefore N=\alpha \\ & P=N \cdot e^{k t}\end{aligned}$
Given P = 2N when t = 60 years,
$\begin{aligned} & \therefore 2 N=N e^{60 k} \\ & \therefore 2=e^{60 k} \Rightarrow k=\frac{1}{60} \log 2 \\ & \therefore P=N \cdot e^{60 k}\end{aligned}$
Required t when P = 3N
$\begin{aligned} & 3=e^{k t} \Rightarrow \log 3=k t \\ & \log 3=\left(\frac{1}{60} \log 2\right) \cdot t \\ & t=\frac{60 \log 3}{\log 2} \\ & =\frac{60 \times 1.0986}{0.6912} \\ & =95.4 \text { years }(\approx .)\end{aligned}$
The population of the countr will triple approximately in 95.4 years.

Given,
$\frac{d y}{d x}=\cos (x+y) \ldots$ (i)
Put $x+y=v$...(ii)
$\therefore y=v-x$
$\therefore \frac{d y}{d x}=\frac{d v}{d x}-1 \ldots$..(iii)
Substituting (ii) and (iii) in (i), we get
$\begin{aligned} & \frac{d v}{d x}-1=\cos v \\ & \therefore \frac{d v}{d x}=1+\cos v \\ & \therefore \frac{d v}{d x}=2 \cos ^2\left(\frac{v}{2}\right) \\ & \therefore \frac{1}{\cos ^2\left(\frac{v}{2}\right)} d v=2 d x \\ & \therefore \sec ^2\left(\frac{v}{2}\right) d v=2 d x\end{aligned}$
Integrating on both sides, we get
$\begin{aligned} & \int \sec ^2\left(\frac{v}{2}\right) d v=2 \int d x \\ & \therefore 2 \tan \left(\frac{v}{2}\right)=2 x+c \prime \\ & \therefore \tan \left(\frac{v}{2}\right)=x+\frac{c \prime}{2} \\ & \therefore \tan \left(\frac{x+y}{2}\right)=x+c, \text { where } c=\frac{c \prime}{2}\end{aligned}$

$\frac{d y}{d x}=\frac{y+\sqrt{x^2+y^2}}{x}$
Put y = vx
$\therefore \frac{d y}{d x}=v+x \frac{d v}{d x}$
$\therefore$ (1)becomes, $v+x \frac{d v}{d x}=\frac{v x+\sqrt{x^2+v^2 x^2}}{x}$
$\therefore v+x \frac{d v}{d x}=v+\sqrt{1+v^2}$
$\therefore \frac{1}{\sqrt{1+v^2}} d v=\frac{1}{x} d x$
Integrating, we get,
$\begin{aligned} & \int \frac{1}{\sqrt{1+v^2}} d v=\int \frac{1}{x} d x+c_1 \\ & \therefore \log \left|v+\sqrt{1+v^2}\right|=\log |x|+\log c, \text { where }_1=\log c \\ & \therefore \frac{y+\sqrt{x^2+y^2}}{x}=c x \\ & \left(y+\sqrt{x^2+y^2}\right)=c x^2 \text { is the general solution }\end{aligned}$

y = A cos (log x) + B sin (log x)    ...(1)
Differentiating w.r.t. x, we get
$\begin{aligned} & \frac{d y}{d x}=-A \sin (\log x) \cdot \frac{d}{d x}(\log x)+B \cos (\log x) \cdot \frac{d}{d x}(\log x) \\ & =\frac{-A \sin (\log x)}{x}+\frac{B \cos (\log x)}{x} \\ & \therefore x \frac{d y}{d x}=-A \sin (\log x)+B \cos (\log x)\end{aligned}$
Differentiating again w.r.t. x, we get
$\begin{aligned} & x \frac{d^2 y}{d x^2}+\frac{d y}{d x}=\frac{-A \cos (\log x)}{x}+\frac{B \sin (\log x)}{x} \\ & \therefore x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}=-[A \cos (\log x)+B \sin (\log x)]=-y \quad \ldots .[\text { By (1)] } \\ & \therefore x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}+y=0 \text { is the required D.E. }\end{aligned}$