Differential equations practice · Medium

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

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

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:

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 $x=f(t) \cos t-f^{\prime}(t) \sin t$ 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 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

A solution of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{x y\left[x^2 \sin y^2+1\right]}$ is
( $C$ is an arbitrary constant)

If, $x = A\cos 4t + B\sin 4t$ then $\frac{{{d^2}x}}{{d{t^2}}}$ is equal to

The solution of the differential equation $\frac{d y}{d x}=\frac{y \cos x-y^2}{\sin x}$ is equal to (where ${c}$ is an arbitrary constant)

If $\sin \left(\frac{y}{x}\right)=\log _e|x|+\frac{\alpha}{2}$ is the solution of the differential equation $x \cos \left(\frac{y}{x}\right) \frac{d y}{d x}=y \cos \left(\frac{y}{x}\right)+x$ and $y(1)=\frac{\pi}{3}$, then $\alpha^2$ is equal to

$\left(x^2+y^2\right) d y=x y d x$. If $y\left(x_0\right)=e, y(1)=1$, then value of $x_0=$

The number of solutions of $y' = \frac{{y + 1}}{{x - 1}},y\left( 1 \right) = 2$ is

If $y=y(x)$ is a function of $x$ and $y(1)=0$, then the solution of the equation $x \frac{d y}{d x}-\frac{y}{x+1}=x$ is

Solution of the equation $x \sin \left(\frac{y}{x}\right) d y=\left(y \sin \frac{y}{x}-x\right) d x$ is

If $\frac{d^2 y}{d x^2}=0$, then

If $x\mathrm{ }dy = y\left( {dx + y\mathrm{ }dy} \right),y\left( 1 \right) = 1$ and $y\left( x \right) > 0$, then $y\left( { - 3} \right)$ is equal to

The solution of the differential equation $\left(x^2+y^2\right) d x-5 x y d y=0, y(1)=0$, is:

The integral factor of equation $\left( {{x^2} + 1} \right)\frac{{dy}}{{dx}} + 2xy = {x^2} - 1$ is

Solution of $2 y \sin x \frac{d y}{d x}=2 \sin x \cos x-y^{2} \cos x, x=\frac{\pi}{2}, y=1$ is given by

Solution of $\frac{d y}{d x}=\frac{x(2 \log x+1)}{\sin y+y \cos y}$ when $y=\pi$ at $x=e$ is

The general solution of the differential equation $\frac{{dy}}{{dx}} + \sin \left( {\frac{{x + y}}{2}} \right) = \sin \left( {\frac{{x - y}}{2}} \right)$ is

The general solution of $\frac{{dy}}{{dx}} = \frac{{2x - y}}{{x + 2y}}$ is

The degree and order of the differential equation $y = px + \sqrt[3]{{{a^2}{p^2} + {b^2}}}$, where $p = \frac{{dy}}{{dx}}$, are respectively

The order of differential equation whose general solution is given by $y = {c_1}{e^{2x + {c_2}}} + {c_3}{e^x} + {c_4}\sin \left( {x + {c_5}} \right)$ is

If the solution $y=y(x)$ of the differential equation $\left(x^4+2 x^3+3 x^2+2 x+2\right) d y-\left(2 x^2+2 x+3\right) d x=0$ satisfies $y(-1)=-\frac{\pi}{4}$, then $y(0)$ is equal to:

Solution of the differential equation $\frac{{dy}}{{dx}} + y{\sec ^2}x = \tan x{\sec ^2}x$ is