Vinod Singh and Shweta Pawar Solutions for Chapter: Differential Equations, Exercise 3: Competitive Thinking

If $y\left(t\right)$ is a solution of $\left(1+t\right)\frac{dy}{dt}-ty=1$ and $y\left(0\right)=-1$, then $y\left(1\right)$ is equal to

The solution of $dy=\cos x\left(2-ycosecx\right)dx$, where $y=2$ when $x=\frac{\pi }{2}$, is

Let $y^{\text{'}}\left(x\right)+y\left(x\right)g^{\text{'}}\left(x\right)=g\left(x\right) g^{\text{'}}\left(x\right), y\left(0\right)=0, x\in R$, where $f^{\text{'}}\left(x\right)$ denotes $\frac{d}{dx}\left(f\left(x\right)\right)$ and $g\left(x\right)$ is a given non-constant differentiable function on $R$ with $g\left(0\right)=g\left(2\right)=0$. Then the value of $y\left(2\right)$ is

Let $y=y\left(x\right)$ be the solution of the differential equation $\sin x\frac{dy}{dx}+y \cos x=4x, x\in \left(0,\pi \right)$. If $y \left(\frac{\mathrm{\pi }}{2}\right)=0$, then $y \left(\frac{\mathrm{\pi }}{6}\right)$ is equal to

If $y\left(x\right)$ satisfies the differential equation $y^{\text{'}}-y\tan x=2x\sec x$ and $y\left(0\right)=0$, then

The population of a city increases at the rate $3\%$ per year. If at time $t$ the population of city is $p$, then find the equation of $p$ in time $t$.

Let $I$ be the purchase value of an equipment and $V\left(t\right)$ be the value after it has been used for $t$ years. The value $V\left(t\right)$ depreciates at a rate given by the differential equation $\frac{dV}{dt}=-k\left(T-t\right)$, where $k>0$ is a constant and $T$ is the total life in years of the equipment. Then, the scrap value $V\left(T\right)$ of the equipment is

Let the population of rabbits surviving at a time $t$ be governed by the differential equation $\frac{dp\left(t\right)}{dt}=\frac{1}{2}p\left(t\right)-200$. If $p\left(0\right)=100$, then $p\left(t\right)$ equals