Question

If $y = y(x)$ is a particular solution of $\sqrt{1 - x^2} \frac{dy}{dx} + \frac{2x}{\sqrt{1 - x^2}} y = x$, $y(0) = 1$, then $y\left(\frac{1}{2}\right) =$

• A. $\frac{\sqrt{3}}{2}$
• B. $\frac{1}{4}$
• C. $\frac{1}{2}$
• D. $0$

Hint:

For a first-order linear differential equation $\frac{dy}{dx} + P(x) y = Q(x)$, the integrating factor is $e^{\int P(x) dx}$, and the solution is $y \cdot IF = \int Q(x) \cdot IF dx + C$. Remember to use the initial condition to find the particular solution.

Answer 1

The Correct Option is A

Step 1: Rewrite the differential equation in standard linear form.
Divide the equation by $\sqrt{1 - x^2}$: $$\frac{dy}{dx} + \frac{2x}{1 - x^2} y = \frac{x}{\sqrt{1 - x^2}}$$ Here, $P(x) = \frac{2x}{1 - x^2}$ and $Q(x) = \frac{x}{\sqrt{1 - x^2}}$.
Step 2: Find the integrating factor (IF).
$$IF = e^{\int P(x) dx} = e^{\int \frac{2x}{1 - x^2} dx}$$ Let $u = 1 - x^2$, $du = -2x dx$. $$\int \frac{2x}{1 - x^2} dx = -\int \frac{du}{u} = -\ln|u| = \ln|u^{-1}| = \ln\left|\frac{1}{1 - x^2}\right|$$ $$IF = e^{\ln\left(\frac{1}{1 - x^2}\right)} = \frac{1}{1 - x^2}$$
Step 3: Find the general solution.
$$y \cdot (IF) = \int Q(x) \cdot (IF) dx + C_1$$ $$y \cdot \frac{1}{1 - x^2} = \int \frac{x}{\sqrt{1 - x^2}} \cdot \frac{1}{1 - x^2} dx + C_1 = \int \frac{x}{(1 - x^2)^{3/2}} dx + C_1$$ Let $v = 1 - x^2$, $dv = -2x dx$. $$\int \frac{x}{(1 - x^2)^{3/2}} dx = -\frac{1}{2} \int v^{-3/2} dv = -\frac{1}{2} \frac{v^{-1/2}}{-1/2} = v^{-1/2} = \frac{1}{\sqrt{1 - x^2}}$$ $$y \cdot \frac{1}{1 - x^2} = \frac{1}{\sqrt{1 - x^2}} + C_1$$ $$y = \sqrt{1 - x^2} + C_1 (1 - x^2)$$
Step 4: Apply the initial condition $y(0) = 1$.
$$1 = \sqrt{1 - 0^2} + C_1 (1 - 0^2) \implies 1 = 1 + C_1 \implies C_1 = 0$$
Step 5: Find the particular solution.
$$y(x) = \sqrt{1 - x^2}$$
Step 6: Evaluate $y\left(\frac{1}{2}\right)$.
$$y\left(\frac{1}{2}\right) = \sqrt{1 - \left(\frac{1}{2}\right)^2} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$$ Thus, $y\left(\frac{1}{2}\right) = \boxed{\frac{\sqrt{3}}{2}}$.