Differential equations practice · Medium

$\frac{d x}{d y}=\frac{1+x-y^2}{y}$ and $x(1)=1$, then $5 x(2)$ is equal to

The tangent at any point $P$ on the curve $y=f(x)$ meets the $y$-axis in $B . N$ is the foot of the perpendicular drawn from $P$ onto the $x$-axis. If the rectangle with $O B$ and $O N(O$ is the origin) as adjacent sides is of constant area and equals $a^2$, then the equation of the curve is

$y=a e^{-1 / x}+b$ is a solution of $\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{y}{x^2}$ when

$f(x)+f^{\prime}(x)=\int\limits_{0}^2 f(t) d t$ and $f(0)=e^{-2}$, then the value of $f(2)-2 f(0)$ is:

Solution of the differential equation $\frac{d y}{d x}=x^3 y^3-x y$

Let $y=y(x)$ be the solution curve of the differential equation $\sec y \frac{d y}{d x}+2 x \sin y=x^3 \cos y,$ $y(1)=0.$ Then $y(\sqrt{3})$ is equal to:

The solution of the differential equation $\left(3 x y+y^2\right) d x+\left(x^2+x y\right) d y=0$ is

If $y = y\left( x \right)$ is the solution of the differential equation, $x\frac{{dy}}{{dx}} + 2y = {x^2}$ satisfying $y(1) = 1,$ then $y\left( {\frac{1}{2}} \right)$ is equal to

If $y=y(x), x \in\left(0, \frac{\pi}{2}\right)$ be the solution curve of the differential equation $\left(\sin ^2 2 x\right) \frac{d y}{d x}+\left(8 \sin ^2 2 x+\right.$ $2 \sin 4 x) y=2 e^{-4 x}(2 \sin 2 x+\cos 2 x)$, with $y\left(\frac{\pi}{4}\right)=e^{-\pi}$ then $y\left(\frac{\pi}{6}\right)$ is equal to

General solution of the differential equation $\frac{{dy}}{{dx}} = \frac{{x + y + 1}}{{x + y - 1}}$ is given by

The solution of the equation $x \frac{d y}{d x}+3 y=x$ is

The solution of the differential equation $x{y^2}dy - \left( {{x^3} + {y^3}} \right)dx = 0$ is

If $y = y(x)$ is the solution curve of the differential equation ${x^2}dy + \left( {y - \frac{1}{x}} \right)dx = 0;$ $x > 0\mathrm{ }\,\,\mathrm{and }\,\,y(1) = 1,\,\,\,\mathrm{then}\,\,\mathrm{ }y\left( {\frac{1}{2}} \right)$ is equal to

Solution of $(x y \cos x y+\sin x y) d x+x^2 \cos x y d y=0$ is

Let $y=y(x)$ be the solution of the differential equation $\left(1+y^2\right) e^{\tan x} d x+\cos ^2 x\left(1+e^{2 \tan x}\right) d y=0$, $y(0)=1$. Then $y\left(\frac{\pi}{4}\right)$ is equal to :

The degree and order respectively of the differential equation of all parabolas whose axis is $x$-axis, are:

The solution of $d y=\cos x(2-y \operatorname{cosec} x) d x$ where $y=2$ when $x=\frac{\pi}{2}$ is

If the area of the figure bounded by a curve, the $x$-axis, and two ordinates, one of which is constant, the other variable is equal to the ratio of the cube of the variable ordinate to the variable abscissa, then the curve is

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

The degree of the differential equation corresponding to the family of curves $y = a{\left( {x + a} \right)^2}$, where a is an arbitrary constant is

The solution of the equation $x \frac{d y}{d x}=y-x \tan \left(\frac{y}{x}\right)$ is

Which one of the following is a differential equation of the family of curves $y=A e^{2 x}+B e^{-2 x}$

Solution of the differential equation $\cos x d y=y(\sin x-y) d x, 0 < x < \pi / 2$ is

General solution of the equation $(x \cos x) \frac{d y}{d x}+y(x \sin x+\cos x)=1$ is

Solution of $\frac{{dy}}{{dx}} = {3^{x + y}}$ is

The solution of differential equation $\frac{d y}{d x}+y=1$ is

If $x^y=e^{x-y}$, then $\frac{d y}{d x}=$

The equation of the curve passing through the point $(1, \pi / 4)$ and tangent at any point of which makes angle $\operatorname{Tan}^{-1}\left(\frac{y}{x}-\cos ^2 \frac{y}{x}\right)$ with the $x$-axis, is

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

The solution of differential equation $\left( {\sin x + \cos x} \right)dy + \left( {\cos x - \sin x} \right)dx = 0$ is