Maths Q20 – JEE-MAIN 2025

Home Maths Differential EquationsQuestion #186

Let

y = y (x)

be the solution curve of the differential equation

x (x^{2} + e^{x}) d y + (e^{x} (x - 2) y - x^{3}) d x = 0, x > 0

, passing through the point

(1, 0)

. Then

y (2)

is equal to

Clue 1 of 3: Recognize the Differential Equation Type

The given differential equation is a first-order equation that can be transformed into an exact differential equation.

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Ninja Strategy•Sign Analysis from Initial Condition

Evaluate the slope at the initial point $(1, 0)$ to determine the expected sign of $y (2)$ , then eliminate options with the wrong sign.

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Step 1: Identify and Transform to Exact Form✦ Active

The given differential equation is $x (x^{2} + e^{x}) d y + (e^{x} (x - 2) y - x^{3}) d x = 0$ . This is of the form $M d x + N d y = 0$ , where $M = e^{x} (x - 2) y - x^{3}$ and $N = x (x^{2} + e^{x})$ . We calculate the partial derivatives:

\frac{\partial M}{\partial y} = e^{x} (x - 2)

\frac{\partial N}{\partial x} = (x^{2} + e^{x}) + x (2 x + e^{x}) = 3 x^{2} + e^{x} + x e^{x}

Since $\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$ , the equation is not exact. We calculate the integrating factor using the formula $\frac{1}{N} (\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x})$ :

\frac{1}{x (x^{2} + e^{x})} (e^{x} (x - 2) - (3 x^{2} + e^{x} + x e^{x})) = \frac{x e^{x} - 2 e^{x} - 3 x^{2} - e^{x} - x e^{x}}{x (x^{2} + e^{x})} = \frac{- 3 x^{2} - 3 e^{x}}{x (x^{2} + e^{x})} = \frac{- 3 (x^{2} + e^{x})}{x (x^{2} + e^{x})} = - \frac{3}{x}

The integrating factor is $I . F . = e^{\int - \frac{3}{x} d x} = e^{- 3 \ln x} = e^{\ln x^{- 3}} = x^{- 3} = \frac{1}{x^{3}}$ . Multiplying the original equation by $x^{- 3}$ yields the exact differential equation:

(\frac{e^{x} (x - 2) y}{x^{3}} - 1) d x + (\frac{x^{2} + e^{x}}{x^{2}}) d y = 0

Let $M^{'} = \frac{e^{x} (x - 2) y}{x^{3}} - 1$ and $N^{'} = \frac{x^{2} + e^{x}}{x^{2}}$ .

Step 2: Solve the Exact Differential Equation○ Expand

The general solution $F (x, y) = C$ is found by integrating $N^{'}$ with respect to $y$ and adding an arbitrary function of $x$ , $g (x)$ :

F (x, y) = \int N^{'} d y + g (x) = \int (1 + \frac{e^{x}}{x^{2}}) d y + g (x) = y (1 + \frac{e^{x}}{x^{2}}) + g (x)

Now, differentiate $F (x, y)$ with respect to $x$ and equate it to $M^{'}$ :

\frac{\partial F}{\partial x} = y \frac{\partial}{\partial x} (\frac{x^{2} + e^{x}}{x^{2}}) + g^{'} (x) = y \frac{(2 x + e^{x}) x^{2} - (x^{2} + e^{x}) (2 x)}{x^{4}} + g^{'} (x)

= y \frac{2 x^{3} + x^{2} e^{x} - 2 x^{3} - 2 x e^{x}}{x^{4}} + g^{'} (x) = y \frac{x^{2} e^{x} - 2 x e^{x}}{x^{4}} + g^{'} (x)

= y \frac{e^{x} (x - 2)}{x^{3}} + g^{'} (x)

Equating this to $M^{'}$ :

y \frac{e^{x} (x - 2)}{x^{3}} + g^{'} (x) = \frac{e^{x} (x - 2) y}{x^{3}} - 1

This implies $g^{'} (x) = - 1$ , so integrating with respect to $x$ gives $g (x) = - x$ . The general solution is:

y (1 + \frac{e^{x}}{x^{2}}) - x = C

Step 3: Apply Initial Condition and Find y(2)○ Expand

The curve passes through the point $(1, 0)$ . Substitute $x = 1, y = 0$ into the general solution to find the constant $C$ :

0 (1 + \frac{e^{1}}{1^{2}}) - 1 = C ⟹ C = - 1

The particular solution is $y (\frac{x^{2} + e^{x}}{x^{2}}) - x = - 1$ . To find $y (2)$ , substitute $x = 2$ :

y (2) (\frac{2^{2} + e^{2}}{2^{2}}) - 2 = - 1

y (2) (\frac{4 + e^{2}}{4}) = 1

y (2) = \frac{4}{4 + e^{2}}

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Maths Question 20 – JEE-MAIN 2025

Clue 1 of 3: Recognize the Differential Equation Type

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