Differential equations practice · Easy

The differential equation $\frac{d y}{d x}=\frac{x+y-1}{x+y+1}$ reduces to variable separable form by making the substitution

A particle starts at the origin and moves along the $x$-axis in such a way that its velocity at the point $\left( {x,0} \right)$ is given by the formula $\frac{{dx}}{{dt}} = {\cos ^2}\pi x$. Then, the particle never reaches the point on

If $x \sin y=\sin (y+a)$ and $\frac{d y}{d x}=\frac{A}{1+x^2-2 x \cos a}$ then the value of $A$ is

The differential equation whose solution is $(x-h)^2+(y-k)^2=a^2$ is ( $a$ is constant)

The solution of the differential equation $\frac{{dy}}{{dx}} = \sin \left( {x + y} \right)\tan \left( {x + y} \right) - 1$ is

If $y=y(x)$ is the solution of the differential equation $\left(1+e^{2 x}\right) \frac{d y}{d x}+2\left(1+y^2\right) e^x=0$ and $y(0)=0$, then $6\left(y^{\prime}(0)+\left(y\left(\log _e \sqrt{3}\right)\right)^2\right)$ is equal to

The differential equation of the family of non-vertical lines passing through the origin is

Solution of $y\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2+2 x \frac{\mathrm{d} y}{\mathrm{~d} x}=y$ is

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

Integrating factor of the differential equation $\frac{d y}{d x}+y \tan x-\sec x=0$ is

If $y=y(x)$ is the solution of the differential equation $\left(1+e^{2 x}\right) \frac{d y}{d x}+2\left(1+y^2\right) e^x=0$ and $y(0)=0$ then which of the following is correct

If $y(x-y)^2=x$, then
$\int \frac{d x}{(x-3 y)}=\frac{m}{n} \ln \left[(x-y)^2-1\right]$.Then $(m+2 n)$ is equal to

The population $p(t)$ at time $t$ of a certain mouse species follows the differential equation $\frac{d p(t)}{d t}=0.5$ $p(t)-450$. If $\mathrm{p}(0)=850$, then the time at which the population becomes zero is

Which of the following equation is non-linear

Let $y=y(x)$ be the solution curve of the differential equation $\frac{d y}{d x}=\frac{y}{x}\left(1+x y^{2}\left(1+\log _{e} x\right)\right), x>0, y(1)=3$. Then $\frac{y^{2}(x)}{9}$ is equal to

Solution of $\frac{d y}{d x}+\sin \left(\frac{x+y}{2}\right)=\sin \left(\frac{x-y}{2}\right)$ is

${\tan ^{ - 1}}x + {\tan ^{ - 1}}y = C$ is the general solution of the differential equation

The solution of $\frac{{dy}}{{dx}} = {\left( {\frac{y}{x}} \right)^{1/3}}$ is

Solution of the differential equation $y d x+(x+y) d y=0$ is

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

The differential equation of the family of curves $y = a\cos \left( {x + b} \right)$ is

Let $y=y(x)$ satisfies the equation $\frac{d y}{d x}-|A|=0$ for $x>0$, where $A=\left[\begin{array}{ccc}y & \sin x & 1 \\ 0 & -1 & 1 \\ 2 & 0 & \frac{1}{x}\end{array}\right]$.

If $y(\pi)=\pi+2$ then $y\left(\frac{\pi}{2}\right)=$

Consider the differential equation:
$\frac{d y}{d x}=\frac{y^3}{2\left(x y^2-x^2\right)}$ Statement-1: The substitution $z=y^2$ transforms the above equation into a first order homogenous differential equation.
Statement-2: The solution of this differential equation is $y^2 e^{-y^2} / x=C$

The differential equation for the family of curves $x^2-y^2-2 a y=0$, where $a$ is an arbitrary constant, is

The curve satisfying $y \mathrm{~d} x-x \mathrm{~d} y+\log x \mathrm{~d} x=0$ $(x > 0)$ passing through $(1,-1)$ is

The solution of differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{y}{x}+2 \frac{\phi(y / x)}{\phi^{\prime}(y / x)}$ is

The solution of the differential equation $\frac{d y}{d x}=-\left(\frac{x^2+3 y^2}{3 x^2+y^2}\right), y(1)=0$

Solution of differential equation $2 x y \frac{d y}{d x}=x^2+3 y^2$ is

The differential equation of the family of curves $y=e^x(A \cos x+B \sin x)$ is

Solution of the equation $x \frac{d y}{d x}-y=\sqrt{x^2+y^2}$ is