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JEE Mathematics 2024 Question with Solution

The differential equation, whose general solution is y=c1f(x)+c2y = c_1 f(x) + c_2, where c1c_1 and c2c_2 are arbitrary constants, is:

  • A

    (8ex−1)d2ydx2+dydx=0(8e^x - 1) \frac{d^2y}{dx^2} + \frac{dy}{dx} = 0

  • B

    (8ex+1)d2ydx2−dydx=0(8e^x + 1) \frac{d^2y}{dx^2} - \frac{dy}{dx} = 0

  • C

    (8ex+1)d2ydx2+dydx=0(8e^x + 1) \frac{d^2y}{dx^2} + \frac{dy}{dx} = 0

  • D

    (8ex−1)d2ydx2−dydx=0(8e^x - 1) \frac{d^2y}{dx^2} - \frac{dy}{dx} = 0

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The general solution is y=c1f(x)+c2y = c_1 f(x) + c_2.

Also, from the solution data,

∫0af(x) dx=e−a+4a2+a−1\int_0^a f(x) \, dx = e^{-a} + 4a^2 + a - 1

Find: The differential equation satisfied by yy.

Differentiate the given integral with respect to aa:

f(a)=dda(e−a+4a2+a−1)=−e−a+8a+1f(a) = \frac{d}{da}\left(e^{-a} + 4a^2 + a - 1\right) = -e^{-a} + 8a + 1

So,

f(x)=−e−x+8x+1f(x) = -e^{-x} + 8x + 1

Now differentiate:

f′(x)=e−x+8f'(x) = e^{-x} + 8

and

f′′(x)=−e−xf''(x) = -e^{-x}

Since

y=c1f(x)+c2y = c_1 f(x) + c_2

we get

dydx=c1f′(x)\frac{dy}{dx} = c_1 f'(x)

and

d2ydx2=c1f′′(x)\frac{d^2y}{dx^2} = c_1 f''(x)

Substitute f′(x)=e−x+8f'(x) = e^{-x} + 8 and f′′(x)=−e−xf''(x) = -e^{-x} into the expression

(8ex+1)d2ydx2+dydx(8e^x + 1)\frac{d^2y}{dx^2} + \frac{dy}{dx}

Then

(8ex+1)c1f′′(x)+c1f′(x)(8e^x + 1) c_1 f''(x) + c_1 f'(x) =c1[(8ex+1)(−e−x)+(e−x+8)]= c_1 \left[(8e^x + 1)(-e^{-x}) + (e^{-x} + 8)\right] =c1[−8−e−x+e−x+8]=0= c_1 \left[-8 - e^{-x} + e^{-x} + 8\right] = 0

Therefore, the required differential equation is

(8ex+1)d2ydx2+dydx=0(8e^x + 1)\frac{d^2y}{dx^2} + \frac{dy}{dx} = 0

Hence, the correct option is C.

Differentiate the given integral first

Given:

∫0af(x) dx=e−a+4a2+a−1\int_0^a f(x) \, dx = e^{-a} + 4a^2 + a - 1

and the family of solutions is y=c1f(x)+c2y = c_1 f(x) + c_2.

Find: Which option gives the correct differential equation.

Using the Fundamental Theorem of Calculus,

f(a)=dda∫0af(x) dxf(a) = \frac{d}{da} \int_0^a f(x) \, dx

So,

f(a)=dda(e−a+4a2+a−1)=−e−a+8a+1f(a) = \frac{d}{da}(e^{-a} + 4a^2 + a - 1) = -e^{-a} + 8a + 1

Replacing aa by xx,

f(x)=−e−x+8x+1f(x) = -e^{-x} + 8x + 1

Then,

f′(x)=e−x+8,f'(x) = e^{-x} + 8, f′′(x)=−e−xf''(x) = -e^{-x}

Now for

y=c1f(x)+c2,y = c_1 f(x) + c_2,

we have

y′=c1f′(x),y′′=c1f′′(x)y' = c_1 f'(x), \qquad y'' = c_1 f''(x)

Test option C:

(8ex+1)y′′+y′=0(8e^x + 1)y'' + y' = 0

Substituting,

(8ex+1)c1f′′(x)+c1f′(x)=0(8e^x + 1)c_1 f''(x) + c_1 f'(x) = 0 c1[(8ex+1)(−e−x)+(e−x+8)]=0c_1\left[(8e^x + 1)(-e^{-x}) + (e^{-x} + 8)\right] = 0 c1(−8−e−x+e−x+8)=0c_1(-8 - e^{-x} + e^{-x} + 8) = 0 0=00 = 0

Thus the family y=c1f(x)+c2y = c_1 f(x) + c_2 satisfies option C. Therefore, the correct differential equation is (8ex+1)d2ydx2+dydx=0(8e^x + 1)\frac{d^2y}{dx^2} + \frac{dy}{dx} = 0.

Common mistakes

  • Differentiating ∫0af(x) dx\int_0^a f(x) \, dx incorrectly with respect to aa. By the Fundamental Theorem of Calculus, it becomes f(a)f(a), not an integral again. First obtain f(a)f(a) correctly, then proceed.

  • Forgetting that c2c_2 is a constant. When differentiating y=c1f(x)+c2y = c_1 f(x) + c_2, the derivative of c2c_2 is 00. So y′=c1f′(x)y' = c_1 f'(x) and y′′=c1f′′(x)y'' = c_1 f''(x).

  • Making a sign error while differentiating −e−x-e^{-x} or e−xe^{-x}. Since ddx(e−x)=−e−x\frac{d}{dx}(e^{-x}) = -e^{-x}, it follows that ddx(−e−x)=e−x\frac{d}{dx}(-e^{-x}) = e^{-x}. Track the negative sign carefully.

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