Maths Solve the Following Question.(4 Marks)

Cos ( x + y )dy = dx
$\therefore \frac{d y}{d x}=\frac{1}{\cos (x+y)}$
Let x + y = t
$\begin{aligned} & \therefore 1+\frac{d y}{d x}=\frac{d t}{d x} \\ & \therefore \frac{d t}{d x}-1=\frac{1}{\cos t} \\ & \frac{d t}{d x}=\frac{1}{\cos t}+1 \\ & \frac{d t}{d x}=\frac{1+\cos t}{\cos t} \\ & \therefore \frac{\cos t}{1+\cos t} d t=d x\end{aligned}$
Integrating both side.
$\begin{aligned} & \therefore \int \frac{\cos t}{1+\cos t} d t=\int d x \\ & \therefore \int \frac{\cos t(1-\cos t)}{\sin ^2 t} d t=x+c \\ & \therefore \int\left(\cos e c t \cdot \cot t-\cot ^2 t\right) d t=x+c \\ & \therefore \int\left(\cos e c t \cdot \cot t-\cos e c^2 t+1\right) d t=x+c\end{aligned}$
$\begin{aligned} & \therefore-\operatorname{cosect}+\cot t + t = x + c \\ & \therefore \frac{\cos t}{\sin t}-\frac{1}{\sin t}+t=x+c \\ & -\tan \left[\frac{x+y}{2}\right]+ x + y = x + c \end{aligned}$
$\begin{aligned} & \therefore-\tan \left[\frac{x+y}{2}\right]+ y = c \\ & \text { Putting } x =0, y =0 \\ & \therefore-\tan \left[\frac{0+0}{2}\right]+0= c \\ & \therefore c =0 \\ & \therefore y =\tan \left[\frac{x+y}{2}\right]\end{aligned}$

Let $\theta^{\circ} C$ he the temperature of the body at any time $t$.
Temperature of air is $10^{\circ} C$ i.e $\theta_{\circ}=10$
According to Newton's law of cooling, we have
$\begin{aligned} & \frac{d \theta}{d t} \propto \theta-\theta_{\circ} \\ & \therefore \frac{d \theta}{d t}=-k\left(\theta-\theta_{\circ}\right), k>0 \\ & \therefore \frac{ d \theta}{ dt }=- K (\theta-10) \\ & \therefore \frac{1}{\theta-10} d \theta=-k d t\end{aligned}$
Integrating both sides, we get
$\begin{aligned} & \int \frac{1}{\theta-10} d \theta=- k \int dt \\ & \therefore \log (\theta-10)=- Kt + C \end{aligned}$
When t = 0 , θ = 110° C
∴ log (110 - 10) = - K × 0 + C        ∴ C = log 100
∴ log (θ - 10) = - Kt + log 100
$\log \left(\frac{\theta-10}{100}\right)=-K t$
Also θ = 60° C ; when t = 1
$\begin{aligned} & \therefore \log \left(\frac{60-10}{100}\right)=-K \times 1 \\ & \therefore K=-\log \left(\frac{1}{2}\right)\end{aligned}$
$\therefore \log \left(\frac{\theta-10}{100}\right)= t \log \left(\frac{1}{2}\right)$
when θ = 35° C , then 
$\begin{aligned} & \log \left(\frac{35-10}{100}\right)= t \log \left(\frac{1}{2}\right) \\ & \log \left(\frac{25}{100}\right)= t \log \left(\frac{1}{2}\right) \\ & \log \left(\frac{1}{4}\right)= t \log \left(\frac{1}{2}\right)\end{aligned}$
- log 4 = - t log 2
$t=\frac{\log 4}{\log 2}=2$
The additional time required for body to cool to 35° C = (2 - 1) = 1 hour

 Since dy/dx represents the slope of a tangent at any point (x, y) on a given curve, according to given condition 
$\frac{d y}{d x}=y+2 x$
$\frac{d y}{d x}-y=2 x$ which is in the form of linear differential equation $\frac{d y}{d x}+P y=Q$
where P=-1 ,Q=2x
I. $f=e^{\int P d x}=e^{\int-1 d x}=e^{-x}$
The solution is given by
$\begin{aligned} & y . e^{-x}=\int Q(I F) d x+C \prime \\ & y . e^{-x}=2 \int x \cdot e^{-x} d x+c \\ & \text { Consider } \int x \cdot e^{-x} d x \text {, Integrating by part }\end{aligned}$
$\begin{aligned} & =x \frac{e^{-x}}{-1}-\int\left[\frac{d}{d x} x \int e^{-x} d x\right] d x \\ & =-x e^{-x}-\int \frac{e^{-x}}{-1} d x \\ & =-x e^{-x}-e^{-x} \\ & y e^{-x}=2\left(-x e^{-x}-e^{-x}\right)+c\end{aligned}$
$y=-2 x-2+c e^x$
$2 x+y+2=c e^x$ is the general solution. Since the curve is passing through origin, $x =$0,y = 0
$\begin{aligned} & 0+0+2=c \cdot e^0 \Rightarrow c=2 \\ & 2 x+y+2=2 e^x\end{aligned}$
 is the equation of the required curve.

Let ‘N’ be the number of bacteria at time’t ’
$\begin{aligned} & \therefore \frac{ dN }{d t} \infty N \\ & \frac{ dN }{d t}=k N \\ & \frac{ dN }{N}=k d t\end{aligned}$
Integrating on both sides, we get
$\begin{aligned} & \int \frac{ d N}{N}=k \int d t \\ & \log N=k t+c \\ & w h e n t=0, N=1000 \\ & c=\log 1000 \\ & \log N=k t+\log 1000 \\ & \log \left(\frac{N}{1000}\right)=k t\end{aligned}$
$N=1000 e^{k t}$......(1)
when t = 1 hour, N = 2000
$e^k=2$
$N=1000 \times(2)^t$......from(1)
when $t=2 \frac{1}{2}$ hours, we get
$\begin{aligned} & N=1000 \times(2)^{\frac{5}{2}} \\ & =1000 \times 4 \times \sqrt{2}=4000 \times 1.414 \\ & N=5656\end{aligned}$
Number of bacteria present after $2 \frac{1}{2}$ hours is 5656