If $f(x)$ is a function such that $\int_0^x (1-t)f(t)dt = \int_0^x f(t)dt; f(1) = 1$, then:
$f(x) = -\frac{1}{x} - 3\ln x + 1$
$x^2 + y^2 + 2ax + 2by + c = 0$
Degree of the differential equation is 1
All of these
Step-by-Step Solution
Key Concept: Differentiate the given integral equation with respect to x using Leibniz rule to convert it into a separable differential equation of the form f'(x)/f(x) = g(x), then solve using integration and the initial condition f(1) = 1 to find the explicit form of f(x).
Differentiating the given integral equation $x(1-x)f(x) + \int_0^x (1-t)f(t)dt = xf(x)$ with respect to $x$ yields $\int_0^x (1-t)f(t)dt - x^2f(x) = (1-x)f(x) = x^2f'(x) + f(x)$. This simplifies to $\frac{f'(x)}{f(x)} = \int (\frac{1}{x^2} - \frac{3}{x^3})dx$, giving $\ln f(x) = -\frac{1}{x} - 3\ln x + c$. Using $f(1) = 1$ yields $c = 1$, so $f(x) = e^{-1/x}/x^3$. The solution curves satisfy $x^2 + y^2 + 2ax + 2by + c = 0$, which is a family of circles with arbitrary constants $a, b, c$, corresponding to a third-order differential equation that reduces to first degree upon differentiation.
Correct Answer: 1,3