Differential Equations
Differential Equations
Allen Star Batch
Grade 12
Question:
The solution of the equation $\int_0^x y(t)dt = (x+1)\int_0^x ty(t)dt, x > 0$ as $y = f(x)$ is:
$y = ce^{-1/x}$
$y = \frac{c}{x^3}$
$y = -\frac{1}{2} - c\ln x$
$y = \frac{ce^{-1/x}}{x^3}$
Step-by-Step Solution
Key Concept: Differentiate the integral equation with respect to x using Leibniz rule to convert it into a separable differential equation of the form y'/y = (1-3x)/x², then integrate both sides to find y = (c/x³)e^(-1/x).
Differentiate the given integral equation $x.y(x) + \int_0^x y(t)dt = (x+1)xy(x) + \int_0^x ty(t)dt$ with respect to $x$ to eliminate the integral. This yields $y(x) + xy'(x) + y(x) = (x+1)y(x) + (x+1)xy'(x) + xy(x)$. Simplifying gives $(1-3x)y(x) = x^2y'(x)$, or $ rac{y'(x)}{y(x)} = rac{1}{x^2} - rac{3}{x}$. Integrating both sides: $\ln y(x) = - rac{1}{x} - 3\ln x + \ln c$, which gives $y(x) = rac{ce^{-1/x}}{x^3}$.
Correct Answer: 4