Differential equations practice · Easy

Let $f$ be a differentiable function such that $x^2 f(x)-x=4 \int\limits_{0}^x t f(t) d t, f(1)=\frac{2}{3}$. Then $18 f(3)$ is equal to

If $a$ is an arbitrary constant, then solution of the differential equation $\frac{d y}{d x}+\sqrt{\frac{1-y^2}{1-x^2}}=0$, is

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

If $x = \sin t,y = \cos pt$, then

$\frac{d^3 y}{d x^3}+2\left[1+\frac{d^2 y}{d x^2}\right]=1$ has degree and order is :

A differential equation of the form $\frac{d y}{d x}=F(x, y)$ is said to be homogeneous, if ${F}({x}, {y})$ is a homogeneous function of degree,

If the length of the subnormal at any point of the curve is constant, then the eccentricity of this curve is

Solution of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}=\sin (x+y)+\cos (x+y) \text { is }$

The solution of $\frac{{dy}}{{dx}} + \sqrt {\left( {\frac{{1 - {y^2}}}{{1 - {x^2}}}} \right)} = 0$ is

Assertion : The differential equation

$\frac{d x}{d y}+x=\cos y \text { and } \frac{d x}{d y}+\frac{-2 x}{y}=y^2 e^{-y}$

are first order linear differential equations.

Reason : The differential equation of the form

$\frac{{dx}}{{dy}}+{P}_1 {x}={Q}_1$

where, $P_1$ and $Q_1$ are constants or functions of $y$ only, is called first order linear differential equation.

The degree of the differential equation
$y(x)=1+\frac{d y}{d x}+\frac{1}{1.2}\left(\frac{d y}{d x}\right)^2+\frac{1}{1.2 .3}\left(\frac{d y}{d x}\right)^3+\ldots \ldots$. Is

On putting $\frac{y}{x}=v$ the differential equation $\frac{d y}{d x}=\frac{2 x y-y^2}{2 x y-x^2}$ is transformed to

The solution to the differential equation $(x+1) \frac{d y}{d x}-y=e^{3 x}(x+1)^2$ is

The solution of the differential equation $\frac{{dy}}{{dx}} - \frac{{y + 3x}}{{{{\log }_e}\left( {y + 3x} \right)}} + 3 = 0$ is (where $C$ is a constant of integration)

The solution of the differential equation $y \mathrm{dx}-\left(x+2 y^2\right) \mathrm{dy}$ $=0$ is $x=\mathrm{f}(y)$. If $\mathrm{f}(-1)=1$, then $\mathrm{f}(\mathrm{a})$ is equal to :

The differential equation of all parabolas whose axes are parallel to axis of $x$, is

The differential equation which represents the family of curves $y=C_1 e^{C_2 x}$, where $C_1$ and $C_2$ are arbitrary constants, is

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

To solve first order linear differential equation, we use following steps

I. Write the solution of the given differential equation as

$y(I F)=\int(\mathrm{Q} \times \text { IF }) d x+C$

II. Write the given differential equation in the from $\frac{d y}{d x}+P y=\mathrm{Q}$, where $P$ and $Q$ are constants or functions of $x$ only.

III. Find the integrating factor (IF) $\mathrm{e}^{\int \mathrm{P} \mathrm{dx}}$

The correct order of the above steps is

The order of the differential equation whose solution is $y=a \cos x+b \sin x+c e^{-x}$ is

The general solution of ${y^2}dx + \left( {{x^2} - xy + {y^2}} \right)dy = 0$ is

The solution of $\log \frac{d y}{d x}=3 x+4 y, y(0)=0$ is

Integrating factor of the differential equation $(x+1) \frac{d y}{d x}-y=e^{3 x}(x+1)^2$, is

Let the solution curve of the differential equation $x d y=\left(\sqrt{x^2+y^2}+y\right) d x, x>0$, intersect the line $x=1$ at $y=0$ and the line $x=2$ at $y=\alpha$. Then the value of $\alpha$ is

The differential equation of the family of curves, ${x^2} = 4b(y + b),\,\,b \in R$ is

The solution of $\cos (x+y) d y=d x$, is

The general solution of the differential equation $x \,d y-y \,d x=\sqrt{x^2+y^2} d x$ is

Order and degree of the differential equation $\frac{d^2 y}{d x^2}=\left\{y+\left(\frac{d y}{d x}\right)^2\right\}^{1 / 4}$ are

The differential equation corresponding to the family of curves $y=e^x(a x+b)$ is

Let $y=y(x)$ satisfies the differential equation $\frac{d y}{d x}=x y^2+x-y^2 \sin x-\sin x$, $y(0)=0$, then $y(x)$ is given by