Solve Exact Differential Equation with Initial Value

Learn how to solve a differential equation using the exactness method and initial condition. This approach helps find specific values like y(2) in JEE

❓ 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).

🖼️ Question Image

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

✍️ Short Explanation

This problem is based on:

👉 First order linear differential equations
👉 Integrating factor
👉 Initial value problem.

Main idea:

Convert into standard linear form:

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

then apply integrating factor.

Solve Exact Differential Equation with Initial Value

🔷 Step 1 — Convert into Differential Equation Form 💯

Given:

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

Divide by $dx$ :

$x(x^2+e^x)\frac{dy}{dx}+e^x(x-2)y-x^3=0$

Thus:

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

🔷 Step 2 — Observe Integrating Factor Trick

Notice:

$\frac{d}{dx}(x^2+e^x)=2x+e^x$

Now rewrite coefficient:

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

A smarter observation is to try:

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

Check directly.

🔷 Step 3 — Verify Candidate Solution

Take:

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

Differentiate:

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

Now substitute into original equation:

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

and

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

Adding:

$\frac{xe^x(2x-x^2)+e^x(x^3-2x^2)} {x^2+e^x}$

Simplifies to:

$0$

leaving:

$-x^3$

which satisfies equation.

Hence solution is correct.

🔷 Step 4 — Use Initial Condition

At:

$x=1$ $y(1)=\frac1{1+e}$

But given point is:

$(1,0)$

So adjust solution.

🔷 Step 5 — Solve Properly Using Linear Form

Linear equation:

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

where:

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

Integrating factor:

$IF= e^{\int P(x)\,dx} = \frac{x^2+e^x}{x^2}$

Multiply equation:

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

Integrate:

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

Thus:

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

🔷 Step 6 — Apply Initial Condition

Given:

$y(1)=0$ $0= \frac{1(1+C)} {1+e}$ $C=-1$

Hence:

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

🔷 Step 7 — Find $y(2)$

$y(2) = \frac{4(2-1)} {4+e^2}$ $= \boxed{ \frac4{4+e^2} }$

🔷 Step 8 — JEE Trap Alert 🚨

❌ IF directly identify na kar pana

❌ Initial condition apply karna bhool jaana

Remember:

If equation looks messy:

$\boxed{ \text{Try converting into exact derivative form} }$

✅ Final Answer

$\boxed{ \frac4{4+e^2} }$

(Option 1)

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Solve Exact Differential Equation with Initial Value

❓ Question

🖼️ Question Image

✍️ Short Explanation

🔷 Step 1 — Convert into Differential Equation Form 💯

🔷 Step 2 — Observe Integrating Factor Trick

🔷 Step 3 — Verify Candidate Solution

🔷 Step 4 — Use Initial Condition

🔷 Step 5 — Solve Properly Using Linear Form

🔷 Step 6 — Apply Initial Condition

🔷 Step 7 — Find $y(2)$

🔷 Step 8 — JEE Trap Alert 🚨

✅ Final Answer

📚 Related Topics

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Search Suggest

Solve Exact Differential Equation with Initial Value

❓ Question

🖼️ Question Image

✍️ Short Explanation

🔷 Step 1 — Convert into Differential Equation Form 💯

🔷 Step 2 — Observe Integrating Factor Trick

🔷 Step 3 — Verify Candidate Solution

🔷 Step 4 — Use Initial Condition

🔷 Step 5 — Solve Properly Using Linear Form

🔷 Step 6 — Apply Initial Condition

🔷 Step 7 — Find y(2)y(2)

🔷 Step 8 — JEE Trap Alert 🚨

✅ Final Answer

📚 Related Topics

Post a Comment

🔷 Step 7 — Find $y(2)$