Differential Equations JEE 2024: 1 February Shift 2

What is the answer to the differential equations JEE 2024 question?

The answer is 1. In this differential equations JEE 2024 problem, the limiting condition forces the exponential coefficient to vanish, making the function constant. The source is JEE Main, 1 February 2024, Shift 2, Mathematics. “Hard” is the question bank’s tag, not an official difficulty classification.

The task concerns a differentiable, real-valued function defined on all real numbers, with a nonzero real parameter:

f^{'} (x) = α f (x) + 3, α \in R ⧵ {0} .

Its initial value and limit are:

f (0) = 1, {lim}_{x \to - \infty} f (x) = 1 .

Determine: $f (- \ln 2) .$

The supplied data lists no answer choices, so no option values or letter can be assigned. This is purely algebraic; no figure is needed.

How does the exponential substitution solve the equation?

The exponential substitution cancels the term containing the unknown function. This is a first-order linear, nonhomogeneous equation because of the additive constant 3. The chapter tag “Homogeneous and linear differential equations” does not make it a homogeneous equation of the following type:

y^{'} = F (\frac{y}{x}) .

Set:

f (x) = e^{α x} g (x) .

Apply the product rule, differentiating both factors:

f^{'} (x) = α e^{α x} g (x) + e^{α x} g^{'} (x) .

Substitute into the given equation:

α e^{α x} g (x) + e^{α x} g^{'} (x) = α e^{α x} g (x) + 3 .

Cancel the matching terms on both sides. Since the exponential is never zero, division is valid everywhere: $e^{α x} g^{'} (x) = 3,$ $g^{'} (x) = 3 e^{- α x} .$

Integrate, using the fact that the parameter is nonzero. The negative sign comes from dividing by the exponent’s derivative:

g (x) = 3 \int e^{- α x} d x = - \frac{3}{α} e^{- α x} + C .

Multiply back by the exponential:

f (x) = e^{α x} (- \frac{3}{α} e^{- α x} + C) = - \frac{3}{α} + C e^{α x} .

Keep the integration constant visible. Whether the required limit exists depends on the coefficient as well as the exponent’s sign.

How does the initial condition fix the integration constant?

Substituting zero fixes the integration constant in terms of the unknown parameter. Do this before taking the limit so that the exponential coefficient stays explicit:

1 = - \frac{3}{α} + C e^{0} = - \frac{3}{α} + C .

Rearranging gives: $C = 1 + \frac{3}{α} .$

The family satisfying the differential equation and initial condition is therefore:

f (x) = - \frac{3}{α} + (1 + \frac{3}{α}) e^{α x} .

The parameter is still undetermined. The remaining limiting condition must identify which nonzero parameter values are admissible; it cannot be applied by assuming that every exponential decays at negative infinity.

Why must you check both signs of the parameter?

The positive case is impossible; the negative case works only when the exponential coefficient is zero. The supplied write-up incorrectly treats the exponential as always decaying at negative infinity. Its behaviour depends on the parameter’s sign.

Case 1: the parameter is positive. Then:

α > 0, x \to - \infty ⟹ e^{α x} \to 0 .

The limiting condition would require:

- \frac{3}{α} = 1 ⟹ α = - 3 .

That contradicts the positive-case assumption. Reject this case rather than accepting a value outside its assumed range.

Case 2: the parameter is negative. Now the exponential grows without bound:

α < 0, x \to - \infty ⟹ e^{α x} \to + \infty .

A nonzero coefficient would make the function diverge. A finite limit therefore requires:

1 + \frac{3}{α} = 0 ⟹ α = - 3 .

This value is consistent with the negative case. Substitute into both the constant term and the exponential coefficient:

f (x) = - \frac{3}{- 3} + (1 + \frac{3}{- 3}) e^{- 3 x} = 1 + 0 \cdot e^{- 3 x} = 1 .

The function equals 1 for every finite real argument. Hence:

f (- \ln 2) = 1 .

Check the differential equation, initial value and limiting condition:

f^{'} (x) = 0, α f (x) + 3 = - 3 + 3 = 0,

f (0) = 1, {lim}_{x \to - \infty} f (x) = 1 .

All conditions hold. No assumption about the limit of the derivative is needed.

Where does the incorrect result of negative seven come from?

It comes from replacing a zero coefficient with negative one. In the supplied working, the correct substitution is:

1 + \frac{3}{- 3} = 0,

but replacing this coefficient with negative one produces the false expression:

f_{wrong} (x) = 1 - e^{- 3 x} .

That expression gives:

f_{wrong} (- \ln 2) = 1 - e^{3 \ln 2} = 1 - 8 = - 7 .

The mistake is losing the integration constant’s correct coefficient during parameter substitution, then failing to recheck the conditions. Reject it immediately using the initial value:

f_{wrong} (0) = 1 - 1 = 0 \neq 1 .

It also fails the limiting condition:

{lim}_{x \to - \infty} (1 - e^{- 3 x}) = - \infty .

For the valid solution, the zero-coefficient term is identically zero at every finite argument:

0 \cdot e^{- 3 x} = 0 (x \in R) .

Simplify first, then take the limit; there is no ambiguous zero-times-infinity evaluation. Since the choices are absent, negative seven is an incorrect intermediate result in the supplied working, not a verified wrong option.

Which three practice questions test the same ideas?

Use these original practice questions, not additional verified JEE PYQs, to test the three method decisions:

Question 1: A zero exponential coefficient.
Question 2: A nonzero growing exponential.
Question 3: Linear versus homogeneous substitution methods.

Solve each before reading its answer. Check the initial condition before evaluating the requested argument.

How do I solve Question 1?

The same exponential substitution gives a constant solution because the initial condition makes the exponential coefficient zero. The question is:

h^{'} (x) = 2 h (x) + 4, h (0) = - 2 .

Find: $h (\ln 2) .$

The general solution and initial condition give:

h (x) = - 2 + C e^{2 x}, - 2 = - 2 + C ⟹ C = 0 .

Therefore:

h (\ln 2) = - 2 .

How do I solve Question 2 and check its limit?

The requested value is 9, but the limit is not finite. The question is:

p^{'} (x) = - 3 p (x) + 3, p (0) = 2 .

Find the following value and decide whether the limit at negative infinity is finite: $p (- \ln 2) .$

The solution and initial condition give:

p (x) = 1 + C e^{- 3 x}, 2 = 1 + C ⟹ C = 1 .

Hence:

p (- \ln 2) = 1 + e^{3 \ln 2} = 9,

{lim}_{x \to - \infty} p (x) = + \infty .

The growing exponential has a nonzero coefficient. A finite initial value does not guarantee a finite limit.

Which substitution should I use for Question 3?

Choose ratio substitution here because it cancels the ratio term immediately. This equation is both linear and homogeneous in the ratio-substitution sense:

y^{'} = 1 + \frac{y}{x}, x > 0, y (1) = 0 .

Find: $y (2) .$

Use:

y = v x, y^{'} = v + x v^{'} .

Then:

v + x v^{'} = 1 + v ⟹ v^{'} = \frac{1}{x} .

Integrate and apply the initial condition:

v = \ln x + C, y = x (\ln x + C),

y (1) = 0 ⟹ C = 0,

y = x \ln x, y (2) = 2 \ln 2 .

Continue with Differential Equations Practice Questions JEE: 5 Solved Drills. For equations with an exponential term, write its coefficient explicitly before taking any infinite limit.

Frequently asked questions

What is the answer to the JEE Main 2024 Shift 2 differential equations question?

For the 1 February 2024 Shift 2 question with f'(x) = αf(x) + 3, f(0) = 1 and limit 1 as x approaches negative infinity, the answer is f(-ln 2) = 1. The conditions force α = -3 and f(x) = 1 for every real x.

Which substitution solves f'(x) = αf(x) + 3?

Set f(x) = e^(αx)g(x) to cancel the term containing f(x). This gives g'(x) = 3e^(-αx), so f(x) = -3/α + Ce^(αx) for nonzero α. Applying f(0) = 1 gives C = 1 + 3/α.

Why do we check both signs of α before taking the limit?

As x approaches negative infinity, e^(αx) decays for positive α but grows for negative α. The positive case would require α = -3, contradicting its assumption. In the negative case, a finite limit requires the exponential coefficient to vanish, giving the valid value α = -3.

Why is -7 the wrong answer to this differential equations question?

The result -7 comes from incorrectly replacing the coefficient 1 + 3/(-3) with -1 instead of 0. That produces f(x) = 1 - e^(-3x), which gives f(0) = 0 and diverges to negative infinity as x approaches negative infinity. It therefore fails both the initial value and the limiting condition.