Differential equations practice · Easy

The slope at any point of a curve $y = f\left( x \right)$ is given by $\frac{{dy}}{{dx}} = 3{x^2}$ and it passes through $\left( { - 1,1} \right)$. The equation of the curve is

The solution of the differential equation $\frac{d y}{d x}+P(x) y=0$ is

Let $y = \,y(x)$ be solution of the differential equation ${\log _e}\left( {\frac{{dy}}{{dx}}} \right) = 3x + 4y$, with $y(0) = \,0$.
If $y\left( { - \frac{2}{3}{{\log }_e}2} \right) = \alpha {\log _e}2$, then the value of $\alpha$ is equal to

The trajectories orthogonal to $(2 a-x) y^2=x^3$ is

If the I.F. of the differential equation $\frac{\mathrm{dy}}{\mathrm{dx}}+5 \mathrm{y}=\cos x$ is $\int \mathrm{e}^{\mathrm{Adx}}$, then $\mathrm{A}=$

Integrating factor of differential equation $\cos x \frac{d y}{d x}+y \sin x=1$ is

Let $y=y(x)$ be the solution of the differential equation, $xy' - y = {x^2}(x\cos x + \sin x),x > 0.\,\,\mathrm{ If }\,\,\,y(\pi ) = \pi ,$ then $y''\left( {\frac{\pi }{2}} \right) + y\left( {\frac{\pi }{2}} \right)$ is equal to

Let $y(x)$ be the solution of the differential equation $2{x^2}dy + \left( {{e^y} - 2x} \right)dx = 0,\,\,x > 0$. If $y(e) = 1$, then $y(1)$ is equal to

Order and degree of the differential equation $\frac{d y}{d x}+2 \sqrt{\sin x} \frac{d x}{d y}=7$ respectively

The equation of the curve whose slope is $\frac{{y - 1}}{{{x^2} + x}}$ and which passes through the point (1,0) is

The question given below consist of an assertion $(A)$ and the reason ( $R$ ) use the following key to choose the appropriate answer.

A: The product of the ordinates of the point of tangency to the curve $\left(1+x^2\right) y=2-x$, where the tangent makes equal intercept with coordinate axes is equal to 1 .

$\mathrm{R}$ : Slope of straight line making equal and same sign intercept with coordinate axes is equal to -1

The solution of $\frac{{dy}}{{dx}} = {2^{y - x}}$ is

The solution of the differential equation $\left(x y^4+y\right) d x-x d y=0$, is

If $y=y(x)$ is the solution of the differential equation $\frac{5+e^x}{2+y} \frac{d y}{d x}+e^x=0$ satisfying $y(0)=1$, then a value of $y\left(\log _e 13\right)$ is

The general solution of $y^{\prime}=\frac{y}{x}+\frac{x}{y} \frac{\phi\left(y^2 / x^2\right)}{\phi^{\prime}\left(y^2 / x^2\right)}$ is

The equation of the curve passing through the origin and satisfying the differential equation $\left(1+x^2\right) \frac{d y}{d x}+2 x y=4 x^2$ is

Solution of the differential equation $x \cos (y / x)(y d x+x d y)+y \sin (y / x)(x d y-y d x)=0$ is -

Let $y = y(x)$ be the solution of the differential equation $(x + 1)y' - y = {e^{3x}}{(x + 1)^2},\,$ with $y(0) = \frac{1}{3}.$ Then, the point $x = - \frac{4}{3}$ for the curve $y = y(x)$ is:

The solution of the differential equation $\frac{d^2 y}{d x^2}=e^x\sin x$ when $y_1\left(\frac{\pi}{4}\right)=0, y\left(\frac{\pi}{2}\right)=0$ is

Let the function $x(t) \& y(t)$ satisfy the differential equation $\frac{d x}{d t}+a x=0$, $\frac{d y}{d t}+b y=0$. If $x(0)=2, y(0)=1$ and $\frac{x(1)}{y(1)}=\frac{3}{2}$ then $x(t)=y(t)$ for $t=$

The solution of $y\mathrm{ }dx - x\mathrm{ }dy + 3{x^2}{y^2}{e^{{x^3}}}dx = 0$ is

The integrating factor of the differential equation $\frac{d y}{d x}-1-e^x y$ is

If a curve $y = f\left( x \right),$ passing through the point (1,2), is the solution of the differential equation, $2{x^2}dy = \left( {2xy + {y^2}} \right)dx,$ then $f\left( {\frac{1}{2}} \right)$ is equal to

The degree of the differential equation $\left(\frac{d^3 y}{d x^3}\right)^4+\left(\frac{d^2 y}{d x^2}\right)^5+\frac{d y}{d x}+y=0$ is

The differential equation $\frac{{{d^2}y}}{{d{x^2}}} = 2$ represents

The order and degree of the differential equation ${\left( {1 + 4\frac{{dy}}{{dx}}} \right)^{\frac{2}{3}}} = 4\frac{{{d^2}y}}{{d{x^2}}}$ are respectively

The degree of the differential equations $x = 1 + \left( {\frac{{dy}}{{dx}}} \right) + \frac{1}{{2!}}{\left( {\frac{{dy}}{{dx}}} \right)^2} + \frac{1}{{3!}}{\left( {\frac{{dy}}{{dx}}} \right)^3} + \ldots$

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

Let $y=f(x)$ be the solution of the differential equation $y(x+1) d x-x^{2} d y=0, y(1)=e$. Then $\lim\limits_{x \to 0} f(x)$ is equal to

The solution curve of the differential equation $\left(1+e^{-x}\right)\left(1+y^2\right) \frac{d y}{d x}=y^2$, which passes through th point $(0,1)$ is