Differential Equations Question - None

Question

Given $(x^2+1)^2 \dfrac{dy}{dx} + 2x(x^2+1)y = 1$ with $y(0) = 0$. If $\sqrt{a}\, y(1) = \dfrac{\pi}{32}$, find the value of $a$.

Accepted Answer

Step 1: Rewrite the equation in standard form by dividing by (x²+1)²:

$$\frac{dy}{dx} + \frac{2x}{x^2+1}y = \frac{1}{(x^2+1)^2}$$

Step 2: Identify the integrating factor. Here P(x) = 2x/(x²+1), so:

$$\mu(x) = e^{\int \frac{2x}{x^2+1}dx} = e^{\ln(x^2+1)} = x^2+1$$

Step 3: Multiply throughout by μ(x) = x²+1:

$$(x^2+1)\frac{dy}{dx} + 2xy = \frac{1}{x^2+1}$$

The left side is $\frac{d}{dx}[(x^2+1)y]$, so:

$$\frac{d}{dx}[(x^2+1)y] = \frac{1}{x^2+1}$$

Step 4: Integrate both sides:

$$(x^2+1)y = \int \frac{1}{x^2+1}dx = \arctan(x) + C$$

Step 5: Apply initial condition y(0) = 0:

$$(0+1)(0) = \arctan(0) + C \implies C = 0$$

Step 6: Therefore: $y = \frac{\arctan(x)}{x^2+1}$

Step 7: Find y(1):

$$y(1) = \frac{\arctan(1)}{1^2+1} = \frac{\pi/4}{2} = \frac{\pi}{8}$$

Step 8: Use the given condition $\sqrt{a}\,y(1) = \frac{\pi}{32}$:

$$\sqrt{a} \cdot \frac{\pi}{8} = \frac{\pi}{32} \implies \sqrt{a} = \frac{1}{4} \implies a = \frac{1}{16} = 0.0625$$

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