Application of Differential Equations

The family of curves that is orthogonal to $xy=c^2$ is -

The orthogonal trajectory of $y^2=bx$ ($b$ being the parameter) is a conic of eccentricity

The orthogonal trajectory of $y^2=4ax$ (where $a$ being parameter) is

The hemispherical tank of radius $2 \mathrm{m}$ is initially full of water and has an outlet of $12{\mathrm{cm}}^2$ cross-sectional area at the bottom. The outlet is opened at some instant. The flow through the outlet is according to the law $v\left(t\right)=0.6\sqrt{2gh\left(t\right)},$ where $v\left(t\right)$ and $h\left(t\right)$ are, respectively, velocity of the flow through the outlet and the height of water level above the outlet at time $t$ and $g$ is the acceleration due to gravity. Find the time it takes to empty the tank.

Find the time required for a cylindrical tank of radius $r$ and height $H$ to empty through a round hole of area $a$ at the bottom. The flow through a hole is according to the law $U\left(t\right)=u\sqrt{2gh\left(t\right)}$, where $v\left(t\right)$ and $h\left(t\right)$ are respectively the velocity of flow through the hole and the height of the water level above the hole at time $t$ and $g$ is the acceleration due to gravity.

A tank initially contains $50$ gallons of fresh water. Brine contains $2$ pounds per gallon of salt, flows into the tank at the rate of $2$ gallons per minute and the mixture kept uniform by stirring runs out at the same rate. If it will take care for the quantity of salt in the tank to increase from $40$ to $80$ pounds (in seconds) is $206\lambda$, then find $\lambda .$

If the population grows at the rate of $5\%$ per year, then the time taken for the population to become double is (Given $\log 2=0·6912$)

The bacteria increases at the rate proportional to the number of bacteria present. If the original number $N_0$ doubles in $4$ hours, then the number of bacteria in $12$ hours will be

The rate of decay of certain substance is directly proportional to the amount present at that instant. Initially, there are $27 \mathrm{gms}$ of certain substance and $3$ hours later it is found that $8 \mathrm{gms}$ are left, then the amount left after one more hour is

For next two question please follow the same

Consider a tank which initially holds $V_0$ liter of brine that contains $a$ lb of salt. Another brine solution, containing $b$ lb of salt per liter is poured into the tank at the rate of $e$ $\mathrm{L}/\min$ while, simultaneously, the well-stirred solution leaves the tank at the rate of $f$ $\mathrm{L}/\min$. The problem is to find the amount of salt in the tank at any time $t$.
Let $Q$ denote the amount of salt in the tank at any time. The time rate of change of $Q$, $\frac{dQ}{dt}$, equals the rate at which salt enters the tank at the rate of $be$ $\mathrm{lb}/\min$. To determine the rate at which salt leaves the tank, we first calculate the volume of brine in the tank at any time $t$, which is the initial volume $V_0$ plus the volume of brine added $et$ minus the volume of brine removed $ft$. Thus, the volume of brine at any time is
$V_0+et-ft ....\left(a\right)$
The concentration of salt in the tank at any time is $Q/\left(V_0+et-ft\right)$ from which it follows that salt leaves the tank at the rate of $f\left(\frac{Q}{V_0+et-ft}\right)\mathrm{lb}/\min$. Thus, 

$\frac{dQ}{dt}=be-f\left(\frac{Q}{V_0+et-ft}\right) .....\left(b\right)$

or $\frac{dQ}{dt}+\frac{f}{V_0+et-ft}Q=be$

 A $50 \mathrm{L}$ tank initially contains $10 \mathrm{L}$ of fresh water. At $t=0$, a brine solution containing $1 \mathrm{lb}$ of salt per gallon is poured into the tank at the rate of $4 \mathrm{L}/\min$ while the well-stirred mixture leaves the tank at the rate of $2 \mathrm{L}/\min$. Then the amount of time required for overflow to occur is

A tangent drawn to the curve $\mathrm{y}=\mathrm{f}\left(\mathrm{x}\right)$ at $\mathrm{P}\left(\mathrm{x}, \mathrm{y}\right)$ cuts the $\mathrm{x} \text{\&} \mathrm{y}$ axis at $\mathrm{A} \text{and} \mathrm{B}$ respectively. If $\mathrm{BP} : \mathrm{AP}=3 : 1$ and $\mathrm{f}\left(1\right)=1$, then the differential equation of curve is

In a bank, principal increases continuously at the rate of $6\%$ per year, the time required to double Rs.$6000$ is

The temperature $T\left(t\right)$ of a body at time $t=0$ is ${160}^{°} F$ and it decreases continuously as per the differential equation $\frac{dT}{dt}=-K\left(T-80\right)$, where $K$ is positive constant. If $T\left(15\right)={120}^{°} F$, then $T\left(45\right)$ is equal to

The curve passing through $P\left({\mathrm{\pi }}^2,\mathrm{}\pi \right)$ is such that for a tangent drawn to it at a point $Q,$ the ratio of the $y$-intercept and the ordinate of $Q$ is $1\mathrm{}:2.$ Then, the equation of the curve is

A gardener is digging a plot of land. As he gets tired, he works more slowly. After '$t$' minutes he is digging at a rate of $\frac{2}{\sqrt{t}}{\mathrm{m}}^2/\min$ . How long will it take him to dig an area of$40 \mathrm{sq} \mathrm{m}$ ?

The orthogonal trajectory of $x^2-y^2=a^2$ , where a is an arbitrary constant, is

The orthogonal trajectories of the family of curves $a^{n-1} y=x^n$ are given by -

The orthogonal trajectories of the family of circles given by $x^2+y^2-2ay=0$ ($a$ is parameter), is

Let $g:R\to R$ be a differentiable function satisfying $g\left(x\right)=g\left(y\right)g\left(x-y\right) \forall x, y\in R$ and $g^{\text{'}}\left(0\right)=a$ and $g^{\text{'}}\left(3\right)=b$, then $g^{\text{'}}\left(-3\right)$ is

A normal at $P\left(x,y\right)$ on a curve meets the $x\mathit{-}$axis at $Q$ and $N$ is the foot of the ordinate at $P$. If $NQ=\frac{x\left(1+y^2\right)}{\left(1+x^2\right)}$  the equation of the curve is, (given that it passes through the point $\left(3,1\right)$)