Find the integral curve of (1-x^2)y'+xy-1=0 that passes…

Find the integral curve of $(1-x^2)y'+xy-1=0$ that passes through the point $(0,1)$.

Step 1: Find the integrating factor

$\mu(x)=e^{\int \frac{x}{1-x^2}dx}=e^{-\frac12\ln|1-x^2|}=(1-x^2)^{-1/2}$.

Step 2: Multiply and integrate

$\dfrac{d}{dx}\!\left[\dfrac{y}{\sqrt{1-x^2}}\right]=\dfrac{1}{(1-x^2)^{3/2}}$. With $x=\sin\theta$, $\displaystyle\int\sec^2\theta\,d\theta=\tan\theta=\dfrac{x}{\sqrt{1-x^2}}$, so $\dfrac{y}{\sqrt{1-x^2}}=\dfrac{x}{\sqrt{1-x^2}}+C$.

Step 3: General solution and initial condition

Hence $y=x+C\sqrt{1-x^2}$. At $(0,1)$: $1=0+C$, so $C=1$.