Question

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

Answer 1

The solution to the differential equation is:

$y = \frac{\sin^{-1}(x) + C}{\sqrt{1-x^2}}$

Answer 2

The given differential equation is a first-order linear differential equation.

1. Rewrite the equation in the standard form $\frac{dy}{dx} + P(x)y = Q(x)$ by dividing by $(1-x^2)$. This yields $P(x) = -\frac{x}{1-x^2}$ and $Q(x) = \frac{1}{1-x^2}$.

2. Calculate the integrating factor $IF = e^{\int P(x) dx}$. The integral $\int -\frac{x}{1-x^2} dx = \frac{1}{2}\ln|1-x^2| = \ln(\sqrt{|1-x^2|})$. Thus, $IF = \sqrt{|1-x^2|}$. Assuming $1-x^2 > 0$, $IF = \sqrt{1-x^2}$.

3. Apply the general solution formula $y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$.

4. Substitute $Q(x)$ and $IF$: $y \sqrt{1-x^2} = \int \frac{1}{1-x^2} \cdot \sqrt{1-x^2} dx + C$.

5. Simplify and evaluate the integral: $\int \frac{1}{\sqrt{1-x^2}} dx = \sin^{-1}(x)$.

6. The solution becomes $y \sqrt{1-x^2} = \sin^{-1}(x) + C$.

7. Solve for $y$: $y = \frac{\sin^{-1}(x) + C}{\sqrt{1-x^2}}$.