The solution of the differential equation _e(dy/dx) = 5x + 2y is given by Option 1: 5e^5x + 2e^(-2y) + C = 0 : C is an arbitrary constant Option 2: 2e^5x + 5e^(-2y) + C = 0 : C is an arbitrary constant Option 3: 2e^(-5x) + 5e^2y + C = 0 : C is an arbitrary constant Option 4: 5e^(-5x) + 2e^2y + C = 0 : C is an arbitrary constant

Answer: 2e^5x + 5e^(-2y) + C = 0 : C is an arbitrary constant. Solution: Given: _e(dy/dx) = 5x + 2y Apply e to both sides: e^(_e(dy/dx)) = e^(5x + 2y) dy/dx = e^(5x + 2y) Using e^(a+b) = e^a * e^b: dy/dx = e^5x * e^2y Separating variables: dy/e^2y = e^5x dx e^(-2y) dy = e^5x dx Integrating both sides: e^(-2y) dy = e^5x dx -1/2e^(-2y) = e^5x/5 + C_1 Multiplying both sides by -10: 5e^(-2y) = -2e^5x - 10C_1 2e^5x + 5e^(-2y) = -10C_1 Since -10C_1 is an arbitrary constant, let C = -10C_1: 2e^5x + 5e^(-2y) + C = 0

Calculus

Differential Equations

Medium

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

$5e^{5x} + 2e^{-2y} + C = 0$ : C is an arbitrary constant

$2e^{5x} + 5e^{-2y} + C = 0$ : C is an arbitrary constant

$2e^{-5x} + 5e^{2y} + C = 0$ : C is an arbitrary constant

$5e^{-5x} + 2e^{2y} + C = 0$ : C is an arbitrary constant

✅ Correct Option: 2

Given:

$\log_e\left(\frac{dy}{dx}\right) = 5x + 2y$

Apply $e$ to both sides:

$e^{\log_e\left(\frac{dy}{dx}\right)} = e^{5x + 2y}$

$\frac{dy}{dx} = e^{5x + 2y}$

Using $e^{a+b} = e^a \cdot e^b$ :

$\frac{dy}{dx} = e^{5x} \cdot e^{2y}$

Separating variables:

$\frac{dy}{e^{2y}} = e^{5x} \, dx$

$e^{-2y} \, dy = e^{5x} \, dx$

Integrating both sides:

$\int e^{-2y} \, dy = \int e^{5x} \, dx$

$-\frac{1}{2}e^{-2y} = \frac{e^{5x}}{5} + C_1$

Multiplying both sides by $-10$ :

$5e^{-2y} = -2e^{5x} - 10C_1$

$2e^{5x} + 5e^{-2y} = -10C_1$

Since $-10C_1$ is an arbitrary constant, let $C = -10C_1$ :

$2e^{5x} + 5e^{-2y} + C = 0$

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Let $y = f(x)$ be the solution curve of the differential equation $\dfrac{dy}{dx} + 2y^2 = 0$ if $f(1) = 1$ , then $f(2)$ is equal to

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Which of the following differential equations are of order 2 and degree 1?

(A). $\frac{d}{dx}\left(\frac{dy}{dx}\right) + x\frac{dy}{dx} = y$

(B). $\frac{d^2y}{dx^2} + x = \sqrt{\frac{dy}{dx} - y}$

(C). $\left(\frac{dy}{dx}\right)^2 + \frac{d^2y}{dx^2} = \log x$

(D). $\left(\frac{dy}{dx}\right)^2 + \frac{dy}{{dx}} = 3$

Choose the correct answer from the options given below:

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

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

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The order and degree of the differential equation $\left[2+\left(\dfrac{dy}{dx}\right)\right]^{5/2}=7\dfrac{d^3y}{dx^3}$ is

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Match List I with List II.

List I	List II
A. $\frac{d^2y}{dx^2} + \frac{dy}{dx} = 0$	I. order 3, degree 1
B. $\left(\frac{d^2y}{dx^2}\right)^2 = 0$	II. order 2, degree 2
C. $\frac{d^3y}{dx^3} + \frac{d^2y}{dx^2} + y = 0$	III. order 2, degree 1
D. $\sin\left(\frac{dy}{dx}\right) + 5y = 0$	IV. order 1, degree is not defined

Choose the correct answer from the options given below:

The solution of the differential equation log⁡e(dydx)=5x+2y\log_e\left(\frac{dy}{dx}\right) = 5x + 2yloge​(dxdy​)=5x+2y is given by

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CUET Mathematics Past Year Questions (Topic-Wise)

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