Differential equations practice · Mixed difficulty

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

Let $y(x)$ be the solution of the differential equation $(x \ln x) \frac{d y}{d x}+y=2 x \ln x,(x \geq 1)$. Then $y(e)$ is equals to:

The solution of the differential equation $x\frac{{dy}}{{dx}} + y = x\cos x + \sin x$, given that $y = 1$ when $x = \frac{\pi }{2}$, is

Consider the following statements

I. The order of the differential equation $\frac{d y}{d x}=e^x$ is 1 .

II. The order of the differential equation $\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}+\mathrm{y}=0$ is 2 .

III. The order of the differential equation

$\left(\frac{d^3 y}{d x^3}\right)+x^2\left(\frac{d^2 y}{d x^2}\right)^3=0 \text { is } 3 .$

Choose correct option.

The differential equation representing the family of ellipses having foci either on the $x$-axis or on the $y$-axis, centre at the origin and passing through the point $(0,3)$ is :

The integrating factor of the differential equation $(x \log x) \frac{d y}{d x}+y=2 \log x$ is

The normal at any point $P(x, y)$ of a curve meets $x$ - and $y$-axes in $A$ and $B$, respectively. If $O$ is the origin and $O A$ and $O B$ are the algebraic intercepts of the normal at $P$ such that $\frac{1}{O A}+\frac{1}{O B}=1$ then the equation of the curve is [given that $(5,4)$ is a point on the curve]

The equation of the curve whose subnormal is twice the abscissa, is

The solution of the differential equation $(x+y-1) d x-(2 x+2 y-3) d y=0$ is $\ldots \ldots$..

The solution of the differential equation $\left(x y^2-e / x^3\right) d x=x^2 y d y ; y(1)=0$, is -

Solution of the differential equation

$x=1+x y \frac{d y}{d x}+\frac{(x y)^2}{2 !}\left(\frac{d y}{d x}\right)^2+\frac{(x y)^3}{3 !}\left(\frac{d y}{d x}\right)^3+\ldots$
is

The order of the differential equation associated with the primitive $y = {c_1} + {c_2}{e^x} + {c_3}{e^{ - 2x + {c_4}}}$, where ${c_1},{c_2},{c_3},{c_4}$ are arbitrary constants, is

Let $\alpha x=\exp \left(x^{\beta} y^{\gamma}\right)$ be the solution of differential equation $2 x^{2} y d y-\left(1-x y^{2}\right) d x=0, x>0$, $y(2)=\sqrt{\log _{e} 2}$. Then $\alpha+\beta-\gamma$ equals

Statement-1: The slope of the tangent at any point $\mathrm{P}$ on a parabola, whose axis is the axis of $x$ and vertex is at the origin, is inversely proportional to the ordinate of the point $P$.
Statement-2: The system of parabolas $y^2=4 a x$ satisfies a differential equation of degree 1 and order 1 .

If $\left(1+x^2\right) \frac{d y}{d x}=x(1-y), y(0)=\frac{4}{3}$, then the value of $y(\sqrt{8})+\frac{8}{9}$ is

The solution of $x^{\prime}+x \tan t=\sec t, x(0)=1$ is

If the function $y = {\sin ^{ - 1}}x$, then $\left( {1 - {x^2}} \right)\frac{{{d^2}y}}{{d{x^2}}}$ is equal to

The differential equation representing the family of curves ${y^2} = 2c\left( {x + \sqrt c } \right)$, where $c > 0$ is a parameter is of order and degree as follows

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

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

If $(2+\sin x) \frac{d y}{d x}+(y+1) \cos x=0$ and $y(0)=1$, then $y\left(\frac{\pi}{2}\right)$ is equal to :

The curve such that the initial ordinate of any tangent is less than the abscissa of the point of tangency by two unit is

If $(2+\sin x) \frac{d y}{d x}+(y+1) \cos x=0$ and $y(0)=1$, then $y\left(\frac{\pi}{2}\right)$ is equal to

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

If $y=y(x)$ is the solution curve of the differential equation $\frac{d y}{d x}+y \tan x=x \sec x, 0 \leq x \leq \frac{\pi}{3}, y(0)=1$, then $y\left(\frac{\pi}{6}\right)$ is equal to

Form of the differential equation of all family of lines $y = mx + \frac{4}{m}$ by eliminating the arbitrary constant $m$ is

If $\frac{d y}{d t}+\alpha \cdot y=\gamma \cdot e^{-\beta t}$, then $\lim\limits_{t \to \infty } y\left( t \right)$ where $\alpha>0, \beta>0, \gamma>0, \alpha \neq \beta$, is equal to

A curve $y=f(x)$ is passing through $(0,0)$. If the slope of the curve at any point $(x, y)$ is equal to $(x+x y)$, then the number of solution of the equation $f(x)=1$, is:

If $c$ is any arbitrary constant, then the general solution of the differential equation $y d x-x d y=x y d x$ is given by

The integrating factor of the differential equation $\frac{{dy}}{{dx}} + y = \frac{{1 + y}}{x}$, is