Differential equations practice · Medium

The solution of differential equation $2 x^3 y d y+\left(1-y^2\right)\left(x^2 y^2+y^2-1\right) d x=0$ is:

If $f(x)$ and $g(x)$ are two solutions of the differential equation $a \frac{d^2 y}{d x^2}+x^2 \frac{d y}{d x}+y=e^x$, then $f(x)-g(x)$ is the solution of

The solution of the differential equation $\frac{d y}{d x}=\frac{1}{x y\left(x^2 \sin y^2+1\right)}$ is

A curve passes through the point $\left( {0,1} \right)$ and the gradient at $\left( {x,y} \right)$ on it is $y\left( {xy - 1} \right)$. The equation of the curve is

The solution of $\left(1+\sin ^2 x\right) d y-\cos x\left(1+y^2\right) d x=0$ given that $y=2$ when, $x=\frac{\pi}{2}$

The solution of the differential equation $\cos y d x+\left(1+2 e^{-x}\right) \sin y d y=0$, where $y(0)=\frac{\pi}{4}$, is

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

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

If $y:[1, \infty) \rightarrow[1, \infty)$ is a function of $\mathrm{x}$ such that $2 x=y^{1 / 5}+y^{-1 / 5}$ and $\left(x^2-1\right) \frac{d^2 y}{d x^2}+x \frac{d y}{d x}=k y(k \in R)$ Then $\mathrm{k}=$

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

The solution of the differential equation $\frac{d y}{d x}+y \cot x=2 \cos x$ is

Solution of the equation ${x^2}y - {x^3}\frac{{dy}}{{dx}} = {y^4}\cos x$, when $y\left( 0 \right) = 1$ is

The solution of the differential equation $\frac{{dy}}{{dx}} = \frac{{x - y + 3}}{{2\left( {x - y} \right) + 5}}$ is

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

The differential equation $\frac{{dy}}{{dx}} = \frac{{x\left( {1 + {y^2}} \right)}}{{y\left( {1 + {x^2}} \right)}}$ represents a family of

If the solution of differential equation $\frac{d y}{d x}=\frac{m x+5}{4 y+f}$ represents a circle, then the values of $m\, \&$ radius $r$ of circle respectively satisfies

The general solution of the differential equation $y d x-x d y-3 x^2 y^2 e^{x^3} d x=0$ is ......

Solution of the differential equation, $\sin y \frac{d y}{d x}=\cos y(1-x \cos y)$ is

Let $y = y(x)$ be the solution of the differential equation $\frac{{dy}}{{dx}} = 2(y + 2\sin x - 5)x - 2\cos x$ such that, $y(0) = 7.$ Then $y(\pi )\,$ is equal to

The differential equation of the family of ellipse having major and minor axes respectively along the $x$ and $y$-axes and the minor axis is equal to half of the major axis, is

For solving $\frac{{dy}}{{dx}} = 4x + y + 1$, suitable substitution is

The particular solution of the differential equation $\frac{d y}{d x}+\frac{2 y}{x}=0$ with initial condition $y(1)=1$ is

Which of the following equation is linear

A curve $y = f\left( x \right)$ passes through the point $P\left( {1,1} \right)$. The normal to the curve at point $P$ is $a\left( {y - 1} \right) + \left( {x - 1} \right) = 0$. If the slope of the tangent at any point on the curve is proportional to the ordinate at that point, then the equation of the curve is

If $y d x+\left(x+x^2 y\right) d y=0$, then $x=$_____

The differential equation $y\frac{{dy}}{{dx}} + x = a$ ($a$ is any constant) represents

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

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