CUET Mathematics: CalculusDifferential Equations. Free, no login required.

The differential equation $y = xp + \sqrt{x^2 p^3 + 4}$ where $p = \frac{dy}{dx}$ is :

(A) of order 1

(B) of degree 1

(C) of order 2

(D) of degree 3

Choose the correct answer from the options given below :

Integerating factor of $(x \log_e x) \frac{dy}{dx} + y = 2 \log_e x$ is :

The equation of curve whose slope is given by $\frac{dy}{dx} = x$ and which passes through $\left(1, \frac{5}{2}\right)$ is :

Match List - I with List - II. Match the integrating factors :

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If m and n are respectively the order and degree of the differential equation :

$\left(\frac{d^2 y}{dx^2}\right)^5 + 6 \frac{\left(\frac{d^2 y}{dx^2}\right)^3}{\frac{d^3 y}{dx^3}} + \frac{d^3 y}{dx^3} = x^2 + 5$, then :

The degree of the differential equation $\left[1 + \left(\frac{dy}{dx}\right)\right]^3 = \left(\frac{d^2 y}{dx^2}\right)^2$ is :

Solution of differential equation $x dy - y dx = 0$ respresents :

The sum of order and degree of the differential equation $\frac{\left\{1+\left(\frac{dy}{dx}\right)^2\right\}^{\frac{5}{2}}}{\frac{d^2y}{dx^2}} = p$ is :

The solution of the differential equation $\frac{dy}{dx} = \frac{6}{x^2}$; $y(1) = 3$ is :

The differential equation whose solution is $Ax^2 + By^2 = 1$ where A and B are arbitrary constant is of :

(A) first order and first degree

(B) second order and first degree

(C) second order and second degree

(D) second order

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Integrating factor of the differential equation $(1 - y^2)\frac{dx}{dy} + xy = ay$ is :

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

If curve represented by differential equation $x \frac{dy}{dx} + y = e^x$ passes through (1, 1), then $y(-1)$ is :

If $U(x) = x + \sqrt{1 + x^2}$, then solution of the differential equation $\frac{dy}{dx} + \sqrt{\frac{1 + y^2}{1 + x^2}} = 0$, is :

The differential equation representing family of curves $y = e^{-2x}(a \cos x + b \sin x)$, where a and b are arbitrary constant, is :

Consider the differential equation $\frac{dy}{dx} = \frac{y+1}{x+1}$, and $y=0$ when $x=2$. The value of $y$ at $x=3$ is :

If the order and the degree of the differential equation $\left(\frac{dy}{dx}\right)^{\frac{1}{2}} = \left(\frac{d^2y}{dx^2}\right)^{\frac{1}{5}}$ are O and S respectively, then $S - O$ is equal to

Match List I with List II

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The curve passing through the point $(-1, 1)$, given that the slope of the tangent to the curve at any point $(x, y)$ is $\frac{2x}{y^2}$ also passes through the point $\left( k, -\frac{1}{2} \right)$, then

If the solution curve of the differentiable equation $\frac{dy}{dx} + 2y = e^{3x}$, passes through the point $\left(0, \frac{6}{5}\right)$, then the value of $y(\log_e 2)$ is:

The sum of order and degree of differential equation $2x^2 \cdot \left(\frac{d^2y}{dx^2}\right) - 3 \cdot \left(\frac{dy}{dx}\right)^3 + y = 0$ is

Match List I with List II

Choose the correct answer from the options given below:

Solution of the differential equation $\frac{dy}{dx} = x + xy - (1 + y)$ is:

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

The solution of differential equation $\sqrt{x+1} - \sqrt{x-1}\frac{dy}{dx} = 0$ is

If the order and degree of the differential equation $\sqrt{\frac{d^2y}{dx^2}} = \left(1 + \frac{dy}{dx}\right)^{\frac{1}{3}}$ are $a$ and $b$ respectively, then the value of $a^2 + b^2$ is

The integrating factor of differential equation $x\frac{dy}{dx} + 2y = x^2 \log x$ is

General solution of the differential equation $\frac{dy}{dx} + \frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}} = 0$ is

A. $\tan^{-1} x + \tan^{-1} y = C$

B. $\sin^{-1} x - \cos^{-1} y = C$

C. $x\sqrt{1 - y^2} - y\sqrt{1 - x^2} = C$

D. $\sin^{-1} x + \sin^{-1} y = C$

E. $\cos^{-1} x + \cos^{-1} y = C$

(where C is arbitrary constant)

Choose the correct answer from the options given below

The differential equation representing family of curves $y = ae^{mx} + be^{nx}$, where $a$ and $b$ are arbitrary constants, is

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