Question

If y = f(x) satisfies the differential equation $(1 + x^2)f'(x) = x(1 - f(x))$, $f(0) = \frac{4}{3}$, then find the value of $f(\sqrt{8}) + \frac{8}{9}$

Answer 1

2

Answer 2

The given differential equation is $(1 + x^2)f'(x) = x(1 - f(x))$. Rearranging the terms, we get: $\frac{f'(x)}{1 - f(x)} = \frac{x}{1 + x^2}$ Integrating both sides: $\int \frac{df}{1 - f} = \int \frac{x}{1 + x^2} dx$ The integral on the left side is $-\ln|1 - f|$. For the integral on the right side, let $u = 1 + x^2$, so $du = 2x dx$. The integral becomes $\int \frac{1}{u} \frac{du}{2} = \frac{1}{2}\ln|u| = \frac{1}{2}\ln(1 + x^2)$. Equating the results: $-\ln|1 - f| = \frac{1}{2}\ln(1 + x^2) + C_1$ Rearranging and exponentiating: $|1 - f| = K (1 + x^2)^{-1/2}$ where $K = e^{-C_1} > 0$. Thus, the general solution is $f(x) = 1 - A (1 + x^2)^{-1/2}$, where $A = \pm K$. Using the initial condition $f(0) = \frac{4}{3}$: $\frac{4}{3} = 1 - A (1 + 0^2)^{-1/2} \implies \frac{4}{3} = 1 - A \implies A = -\frac{1}{3}$ So, the particular solution is: $f(x) = 1 - \left(-\frac{1}{3}\right) (1 + x^2)^{-1/2} = 1 + \frac{1}{3\sqrt{1 + x^2}}$ Now, we find $f(\sqrt{8})$: $f(\sqrt{8}) = 1 + \frac{1}{3\sqrt{1 + (\sqrt{8})^2}} = 1 + \frac{1}{3\sqrt{1 + 8}} = 1 + \frac{1}{3\sqrt{9}} = 1 + \frac{1}{9} = \frac{10}{9}$ Finally, we calculate $f(\sqrt{8}) + \frac{8}{9}$: $f(\sqrt{8}) + \frac{8}{9} = \frac{10}{9} + \frac{8}{9} = \frac{18}{9} = 2$