The differential equation representing the curve y = e^2x(a + bx), where a, b are arbitrary constants is Option 1: d^2y/dx^2 + 4dy/dx + 4y = 0 Option 2: d^2y/dx^2 - 4dy/dx - 4y = 0 Option 3: d^2y/dx^2 - 4dy/dx + 4y = 0 Option 4: d^2y/dx^2 + 4dy/dx - 4y = 0

Answer: d^2y/dx^2 - 4dy/dx + 4y = 0. Solution: We need to eliminate the two arbitrary constants a and b from y = e^2x(a + bx). Since there are 2 constants, we differentiate twice. y = e^2x(a + bx) s (1) Differentiating with respect to x using the product rule: dy/dx = 2e^2x(a + bx) + be^2x Since e^2x(a + bx) = y from equation (1): dy/dx = 2y + be^2x s (2) Differentiating equation (2) with respect to x: d^2y/dx^2 = 2dy/dx + 2be^2x From equation (2), be^2x = dy/dx - 2y. Substituting: d^2y/dx^2 = 2dy/dx + 2(dy/dx - 2y) d^2y/dx^2 = 4dy/dx - 4y Rearranging: d^2y/dx^2 - 4dy/dx + 4y = 0

Calculus

Differential Equations

Medium

applied

The differential equation representing the curve $y = e^{2x}(a + bx)$ , where a, b are arbitrary constants is

$\frac{d^2y}{dx^2} + 4\frac{dy}{dx} + 4y = 0$

$\frac{d^2y}{dx^2} - 4\frac{dy}{dx} - 4y = 0$

$\frac{d^2y}{dx^2} - 4\frac{dy}{dx} + 4y = 0$

$\frac{d^2y}{dx^2} + 4\frac{dy}{dx} - 4y = 0$

✅ Correct Option: 3

We need to eliminate the two arbitrary constants $a$ and $b$ from $y = e^{2x}(a + bx)$ . Since there are 2 constants, we differentiate twice.

$y = e^{2x}(a + bx) \quad \cdots (1)$

Differentiating with respect to $x$ using the product rule:

$\dfrac{dy}{dx} = 2e^{2x}(a + bx) + be^{2x}$

Since $e^{2x}(a + bx) = y$ from equation (1):

$\dfrac{dy}{dx} = 2y + be^{2x} \quad \cdots (2)$

Differentiating equation (2) with respect to $x$ :

$\dfrac{d^2y}{dx^2} = 2\dfrac{dy}{dx} + 2be^{2x}$

From equation (2), $be^{2x} = \dfrac{dy}{dx} - 2y$ . Substituting:

$\dfrac{d^2y}{dx^2} = 2\dfrac{dy}{dx} + 2\left(\dfrac{dy}{dx} - 2y\right)$

$\dfrac{d^2y}{dx^2} = 4\dfrac{dy}{dx} - 4y$

Rearranging:

$\boxed{\dfrac{d^2y}{dx^2} - 4\dfrac{dy}{dx} + 4y = 0}$

2025: 26 May Shift 2

Curd is at 80° F, five minutes later it came down at 60°F. After another 5 minutes, its temperature became 50° F. Given that the rate of change of temperature is proportional to (T - S), where S is temperature of the surroundings and T is temperature of the curd at any time t. Then the temperature of the surroundings is :

2025: 22 May Shift 1

Which of the following statements is/are true?

(A) $(\tan^{-1} y - x)dy = (1 + y^2)dx$ is a differential equation where variables are separable.

(B) $(1 + x^2)dy + 2xydy = \cot x \ dx (x \neq 0)$ is a first order linear differential equation.

(D) $(xy)dx - (x + y^2)dy = 0$ is a homogeneous differential equation.

Choose the correct answer from the options given below:

2022: 10 Aug Shift 1

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

2026: 19 May Shift 1

Match List-I with List-II

List-I (Differential equation)	List-II (Integrating factor)
(A) $\frac{dy}{dx} + 2y = x$	(I) $e^x$
(B) $\frac{dy}{dx} + \left(\frac{2}{x}\right)y = x^2$	(II) $e^{2x}$
(C) $\frac{dy}{dx} - \left(\frac{2}{x}\right)y = \sin x$	(III) $\frac{1}{x^2}$
(D) $\frac{dy}{dx} + y = e^{-x}$	(IV) $x^2$

Choose the correct answer from the options given below:

2022: 6 Aug Shift 2

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:

2026: 29 May Shift 1

Match List-I with List-II

List-I (Differential Equation)	List-II (Degree)
(A) $\left(\dfrac{dy}{dx}\right)^3+\left(\dfrac{d^2y}{dx^2}\right)^2=0$	(I) Not defined
(B) $\left(\dfrac{d^2y}{dx^2}\right)^3+\left(\dfrac{dy}{dx}\right)^2+1=0$	(II) 2
(C) $\sqrt{1+\dfrac{d^2y}{dx^2}}=\dfrac{dy}{dx}+x$	(III) 1
(D) $\dfrac{d^2y}{dx^2}+e^{\frac{dy}{dx}}=0$	(IV) 3

Choose the correct answer from the options given below:

The differential equation representing the curve y=e2x(a+bx)y = e^{2x}(a + bx)y=e2x(a+bx), where a, b are arbitrary constants is

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The differential equation representing the curve $y = e^{2x}(a + bx)$ , where a, b are arbitrary constants is