Let y=y(x) be the solution of the differential equation: dy/dx + (6x^2 + (3x^2+2x^3+4)e^-2x)/((x^3+2)(2+e^-2x))y = 2+e^-2x, x ∈ (-1,2), satisfying y(0) = 3/2. If y(1) = α(2+e^-2), then α is equal to:

The correct answer is Option 4. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q19 – JEE-MAIN 2026

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Home Maths Differential EquationsQuestion #176

Let

y = y (x)

be the solution of the differential equation:

\frac{d y}{d x} + (\frac{6 x^{2} + (3 x^{2} + 2 x^{3} + 4) e^{- 2 x}}{(x^{3} + 2) (2 + e^{- 2 x})}) y = 2 + e^{- 2 x},

x \in (- 1, 2)

, satisfying

y (0) = \frac{3}{2}

. If

y (1) = α (2 + e^{- 2})

, then

α

is equal to:

Clue 1 of 3: Recognize the form

The given differential equation is a first-order linear differential equation of the form $\frac{d y}{d x} + P (x) y = Q (x)$ .

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Step 1: Identify and Simplify

P (x)

✦ Active

The given differential equation is a first-order linear ODE of the form $\frac{d y}{d x} + P (x) y = Q (x)$ . We identify $P (x)$ and simplify it by factoring the numerator:

P (x) = \frac{6 x^{2} + (3 x^{2} + 2 x^{3} + 4) e^{- 2 x}}{(x^{3} + 2) (2 + e^{- 2 x})} = \frac{3 x^{2} (2 + e^{- 2 x}) + 2 e^{- 2 x} (x^{3} + 2)}{(x^{3} + 2) (2 + e^{- 2 x})}

P (x) = \frac{3 x^{2}}{x^{3} + 2} + \frac{2 e^{- 2 x}}{2 + e^{- 2 x}}

Step 2: Calculate Integrating Factor and Solve the ODE○ Expand

Next, we calculate the integrating factor (IF) and solve the differential equation:

\int P (x) d x = \int (\frac{3 x^{2}}{x^{3} + 2} + \frac{2 e^{- 2 x}}{2 + e^{- 2 x}}) d x = \ln (x^{3} + 2) - \ln (2 + e^{- 2 x}) = \ln (\frac{x^{3} + 2}{2 + e^{- 2 x}})

I F = e^{\int P (x) d x} = \frac{x^{3} + 2}{2 + e^{- 2 x}}

The solution is y \cdot I F = \int Q (x) \cdot I F d x

y \cdot \frac{x^{3} + 2}{2 + e^{- 2 x}} = \int (2 + e^{- 2 x}) \cdot \frac{x^{3} + 2}{2 + e^{- 2 x}} d x = \int (x^{3} + 2) d x

y \cdot \frac{x^{3} + 2}{2 + e^{- 2 x}} = \frac{x^{4}}{4} + 2 x + C

💡 Teacher's Secret Hint

Remember to handle the absolute values in the logarithm based on the given domain $x \in (- 1, 2)$ .

Step 3: Apply Initial Condition and Find

α

○ Expand

We use the initial condition $y (0) = \frac{3}{2}$ to find the constant $C$ , then evaluate $y (1)$ to determine $α$ :

Substituting x = 0, y = \frac{3}{2} : \frac{3}{2} \cdot \frac{0^{3} + 2}{2 + e^{0}} = \frac{0^{4}}{4} + 2 (0) + C ⟹ \frac{3}{2} \cdot \frac{2}{3} = C ⟹ C = 1

So, the particular solution is y (x) = \frac{2 + e^{- 2 x}}{x^{3} + 2} (\frac{x^{4}}{4} + 2 x + 1)

For x = 1 : y (1) = \frac{2 + e^{- 2}}{1^{3} + 2} (\frac{1^{4}}{4} + 2 (1) + 1) = \frac{2 + e^{- 2}}{3} (\frac{1}{4} + 3)

y (1) = \frac{2 + e^{- 2}}{3} (\frac{13}{4}) = \frac{13}{12} (2 + e^{- 2})

Comparing with y (1) = α (2 + e^{- 2}), we find α = \frac{13}{12}

💡 Teacher's Secret Hint

Double-check your arithmetic when substituting values and simplifying fractions.

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Maths Question 19 – JEE-MAIN 2026

Clue 1 of 3: Recognize the form

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