Exact Differential Equation Solved in 1 Minute | JEE Main Maths ⚡

 

❓ Question

Let $y = y(x)$ be the solution curve of the differential equation

$$x(x^2 + e^x)\,dy + \big(e^x(x - 2)y - x^3\big)\,dx = 0,\quad x>0,$$

passing through the point $(1,0)$.

Find the value of

$$y(2).$$

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🖼️ Question Image

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✍️ Short Solution

This is a first-order differential equation which is not exact initially.
The correct JEE approach is:

👉 Convert it into a linear DE in $y$
👉 Find a suitable integrating factor (IF)
👉 Use the given point to find the constant
👉 Evaluate $y(2)$

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🔹 Step 1 — Write the DE in standard form

Given:

$$x(x^2 + e^x)\,dy + \big(e^x(x - 2)y - x^3\big)\,dx = 0$$

Rearrange:

$$x(x^2 + e^x)\,dy = \big(x^3 - e^x(x - 2)y\big)\,dx$$

Divide by $x(x^2 + e^x)$:

$$\frac{dy}{dx} + \frac{e^x(x-2)}{x(x^2 + e^x)}\,y = \frac{x^2}{x^2 + e^x}$$

This is a linear differential equation:

$$\frac{dy}{dx} + P(x)y = Q(x)$$

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🔹 Step 2 — Find the Integrating Factor

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

Instead of integrating directly, apply the exact-equation IF shortcut.
The integrating factor turns out to be:

$$\text{IF} = x^{-3}$$

(This is a standard JEE trick — avoids heavy integration.)

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🔹 Step 3 — Multiply the equation by IF

Multiplying entire equation by $x^{-3}$, it becomes exact.

After simplification and integration:

$$\frac{y{e}^x}{x^2}-x+y=C$$

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🔹 Step 4 — Apply the given point $(1,0)$

Substitute $x=1,\; y=0$:

$$0 - 1 + 0 = C \Rightarrow C = -1$$

So the solution curve is:

$$\frac{ye^x}{x^2} + y - x = -1$$

Factor $y$:

$$y\left(1 + \frac{e^x}{x^2}\right) = x - 1$$

Hence,

$$y = \frac{x - 1}{1 + \frac{e^x}{x^2}}$$

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🔹 Step 5 — Find $y(2)$

Substitute $x=2$:

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

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✅ Final Answer

$$\boxed{\dfrac{4}{e^2 + 4}}$$

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