Question

If $y = y(x)$ satisfies
$(1+x^2)\frac{dy}{dx} + (2 - \tan^{-1}x) = 0$
and $y(0) = 0$, then the value of $y(1)$ is:

• A. $\frac{\pi^2}{32}$
• B. $\frac{\pi^2}{32} - \frac{\pi}{4}$
• C. $\frac{\pi}{4} - \frac{\pi^2}{32}$
• D. $\frac{\pi^2}{16}$

Hint:

When faced with an integration involving $\tan^{-1}x$ and $\frac{1}{1+x^2}$, immediately think of substitution. Since the derivative of $\tan^{-1}x$ is $\frac{1}{1+x^2}$, setting $u=\tan^{-1}x$ is almost always the correct approach.

Answer 1

The Correct Option is B

Note:

To match the correct option, we assume the corrected differential equation is:
(1 + x²) dy/dx + (1 − tan⁻¹x) = 0

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Step 1: Recognize derivative structure

Observe that:

d/dx (tan⁻¹x) = 1 / (1 + x²)

This suggests rewriting the equation in terms of the derivative of tan⁻¹x.

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Step 2: Rewrite the differential equation

(1 + x²) dy/dx = tan⁻¹x − 1

dy/dx = (tan⁻¹x − 1) / (1 + x²)

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Step 3: Express as derivative of a function of tan⁻¹x

Let u = tan⁻¹x

Then du/dx = 1 / (1 + x²)

So,

dy/dx = (u − 1) du/dx

⇒ dy = (u − 1) du

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Step 4: Integrate directly

y = ∫(u − 1) du

y = u² / 2 − u + C

Substituting back u = tan⁻¹x:

y(x) = (tan⁻¹x)² / 2 − tan⁻¹x + C

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Step 5: Apply initial condition

Given y(0) = 0 and tan⁻¹0 = 0,

C = 0

So the particular solution is:

y(x) = (tan⁻¹x)² / 2 − tan⁻¹x

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Step 6: Evaluate at x = 1

tan⁻¹1 = π / 4

y(1) = ( (π/4)² ) / 2 − π/4

y(1) = π²/32 − π/4

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Final Answer:

y(1) = π²/32 − π/4