Separable Differential Equation Shortcut

 

❓ Question

If $y=f(x)$ satisfies the differential equation

$$x\frac{dy}{dx}-\sin 2y=x^3\cos^2 y$$

and

$$y(1)=\frac{\pi}{4},$$

then find:

$$y\left(\frac{\pi}{3}\right)$$

Options:

1. $1$
   
2. $\tan^{-1}\left(\frac{\pi}{4}\right)$
   
3. $\tan^{-1}\left(\frac{\pi^3}{27}\right)$
   
4. $0$
   

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🖼 Question Image

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✍️ Short Explanation

This problem is based on:

👉 Differential equations
👉 Trigonometric substitution
👉 Variable separable transformation.

Main idea:

Use:

$$t=\tan y$$

because terms contain:

$$\sin2y,\ \cos^2 y$$

which simplify beautifully.

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🔷 Step 1 — Use Trigonometric Identities 💯

Given:

$$x\frac{dy}{dx}-\sin2y=x^3\cos^2 y$$

Using:

$$\sin2y=2\tan y\cos^2 y$$

Let:

$$t=\tan y$$

Then:

$$\frac{dt}{dx}=\sec^2 y\frac{dy}{dx}$$

and:

$$\frac{dy}{dx}=\cos^2 y\frac{dt}{dx}$$

Substitute into equation:

$$x\cos^2 y\frac{dt}{dx}-2t\cos^2 y=x^3\cos^2 y$$

Cancel $\cos^2 y$:

$$x\frac{dt}{dx}-2t=x^3$$

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🔷 Step 2 — Solve Linear Differential Equation

Rewrite:

$$\frac{dt}{dx}-\frac{2}{x}t=x^2$$

This is linear DE.

Integrating factor:

$$IF=e^{\int -2/x\,dx}$$
$$=x^{-2}$$

Multiply throughout:

$$x^{-2}\frac{dt}{dx}-2x^{-3}t=1$$

LHS becomes:

$$\frac{d}{dx}(t x^{-2})=1$$

Integrate:

$$t x^{-2}=x+C$$

Thus:

$$t=x^3+Cx^2$$

Since:

$$t=\tan y$$

we get:

$$\tan y=x^3+Cx^2$$

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🔷 Step 3 — Apply Initial Condition

Given:

$$y(1)=\frac{\pi}{4}$$

So:

$$\tan\frac{\pi}{4}=1$$

Hence:

$$1=1+C$$
$$C=0$$

Thus:

$$\boxed{ \tan y=x^3 }$$

Therefore:

$$\boxed{ y=\tan^{-1}(x^3) }$$

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🔷 Step 4 — Find $y\left(\frac{\pi}{3}\right)$

Substitute:

$$x=\frac{\pi}{3}$$
$$y\left(\frac{\pi}{3}\right) = \tan^{-1}\left(\left(\frac{\pi}{3}\right)^3\right)$$
$$= \boxed{ \tan^{-1}\left(\frac{\pi^3}{27}\right) }$$

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🔷 JEE Trap Alert 🚨

❌ $\sin2y$ directly expand karke messy algebra kar dena

Best substitution:

$$\boxed{ t=\tan y }$$

because:

$$\sin2y=2\tan y\cos^2 y$$

and:

$$\frac{dy}{dx}=\cos^2 y\frac{dt}{dx}$$

Everything cancels instantly.

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✅ Final Answer

$$\boxed{ \tan^{-1}\left(\frac{\pi^3}{27}\right) }$$

(Option 3)

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