Differential equations practice · Mixed difficulty

Water is drained from a vertical cylindrical tank by opening a valve at the base of the tank. It is known that the rate at which the water level drops is proportional to the square root of water depth $y$, where the constant of proportionality $k>0$ depends on the acceleration due to gravity and the geometry of the hole. Ift is measured in minutes and $k=\frac{1}{15}$, then the time to drain the tank, if the water is $4 \mathrm{~m}$ deep to start with is

If $\frac{{dy}}{{dx}} + \frac{{{2^{x - y}}({2^y} - 1)}}{{{2^x} - 1}} = 0,x,y > 0,y(1) = 1$ then $y(2)$ is equal to

The equation of the curve passing through the origin and satisfying the equation $\left(1+x^2\right) \frac{d y}{d x}+2 x y=4 x^2$ is

Let
$y=f(x)=\sin ^{3}\left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{\frac{3}{2}}\right)\right)$ Then at $x=1$,

The differential equation of all parabolas whose axis of symmetry is along $x$-axis is of order

The solution of $\left(x-y \cos \frac{y}{x}\right) d x+x \cos \frac{y}{x} d y=0$, $y(1)=0$ is given by

The solution of $y_2-7 y_1+12 y=0$ is

The solution of the differential equation $x\frac{{dy}}{{dx}} + 2y = {x^2}(x \ne 0)$ with $y(1) = 1,$ is

The differential equation $\cot y\mathrm{ }dx = x\mathrm{ }dy$ has a solution of the form

The general solution of the differential equation $2 y e^{x / y} d x+\left(y-2 x e^{x / y}\right) d y=0$ is

If $\left(2 x y-y^2-y\right) d x=\left(2 x y+x-x^2\right) d y$ and $y(1)=1$, then the value of $12|y(-1)|$ is

The differential equation of all non-horizontal lines in a plane is

The solution of the differential equation $\frac{d y}{d x}+y=\cos x$ is

The degree of the differential equation satisfying the relation $\sqrt {1 + {x^2}} + \sqrt {1 + {y^2}} = \lambda \left( {x\sqrt {1 + {y^2}} - y\sqrt {1 + {x^2}} } \right)$, is

The solution of the differential equation $\frac{d^2 y}{d x^2}=-\frac{1}{x^2}$ is

Solution of the differential equation $y\left(x y+2 x^2 y^2\right) d x+x\left(x y-x^2 y^2\right) d y=0$ is given by

Let $y = y(x)$ be the solution of the differential equation, $\frac{{dy}}{{dx}} + y\tan x = 2x + {x^2}\tan x,x \in \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)$, such that y(0)=1. Then

The solution of the differential equation $\cos x \sin y \,d x+\sin x \cos y \,d y=0$, is

The solution of $\log \left( {\frac{{dy}}{{dx}}} \right) = ax + by$ is

Consider the differential equation $y^2 d x+$ $\left(x-\frac{1}{y}\right) d y=0$. If $y(1)=1$, then $x$ is given by:

$y=c_1 e^x-c_2 e^{-x}$ is the general solution of a differential equation. The order of the differential equation is , where $c_1$ and $c_2$ are arbitrary constants.

If ${c_1},{c_2},{c_3},{c_4},{c_5}$ and ${c_6}$ are constants, then the order of the differential equation whose general solution is given by
$y = {c_1}\cos (x + {c_2}) + {c_3}\sin (x + {c_4}) + {c_5}{e^x} + {c_6}$ is

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

If $x^2+y^2=1$, then $\left[\right.$ Note $\left.: y^{\prime}=\frac{\mathrm{dy}}{\mathrm{dx}}, \mathrm{y}^{\prime \prime}=\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\right]$

If $x=f(t) \cos t-f^{\prime}(t) \sin t$ and $y=f(t) \sin t+f^{\prime}(t) \cos t$, then $\left(\frac{d x}{d t}\right)^2+\left(\frac{d y}{d t}\right)^2$ is equal to

Let the solution curve $y = y\left( x \right)$ of the differential equation
$\left[ {\frac{x}{{\sqrt {{x^2} - {y^2}} }} + {e^{\frac{y}{x}}}} \right]x\frac{{dy}}{{dx}} = x + \left[ {\frac{x}{{\sqrt {{x^2} - {y^2}} }} + {e^{\frac{y}{x}}}} \right]y$
pass through the points (1, 0) and $\left( {2\alpha ,\alpha } \right),\,\,\alpha > 0.$ Then $\alpha$ is equal to

Let the function $x(t) \& y(t)$ satisfy the differential equation $\frac{d x}{d t}+a x=0$, $\frac{d y}{d t}+b y=0$. If $x(0)=2, y(0)=1$ and $\frac{x(1)}{y(1)}=\frac{3}{2}$ then $x(t)=y(t)$ for $t=$

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

The area enclosed by the closed curve $C$ given by the differential equation $\frac{d y}{d x}+\frac{x+a}{y-2}=0, y(1)=0$ is $4 \pi$.
Let $P$ and $Q$ be the points of intersection of the curve $C$ and the $y$-axis. If normals at $P$ and $Q$ on the curve $C$ intersect $x$-axis at points $R$ and $S$ respectively, then the length of the line segment $R S$ is

If $\frac{d y}{d x}=f(x)+\int_0^1 f(x) d x$, then the equation of the curve $y=f(x)$ passing through $(0,1)$ is