Derive Differential Equation
Given the integral equation for a differentiable function $f: [1, \infty) \rightarrow \mathbb{R}$:
$6\int_1^x f(t)dt = 3x f(x) + x^3 - 4$
Differentiate both sides with respect to $x$:
- LHS derivative: $\frac{d}{dx} \left( 6\int_1^x f(t)dt \right) = 6f(x)$ (using the Fundamental Theorem of Calculus).
- RHS derivative: $\frac{d}{dx} \left( 3x f(x) + x^3 - 4 \right) = \left( 3 \cdot f(x) + 3x \cdot f'(x) \right) + 3x^2 - 0 = 3f(x) + 3xf'(x) + 3x^2$ (using the Product Rule).
Equating the derivatives gives:
$6f(x) = 3f(x) + 3xf'(x) + 3x^2$
Simplify the equation:
$3f(x) = 3xf'(x) + 3x^2$
$f(x) = xf'(x) + x^2$
Rearrange to get the standard form of a differential equation:
$xf'(x) - f(x) = -x^2$
Solve Differential Equation
Divide by $x$ (since $x \ge 1$, $x \neq 0$):
$f'(x) - \frac{1}{x}f(x) = -x$
This is a first-order linear differential equation. Calculate the integrating factor (IF):
$IF = e^{\int -\frac{1}{x}dx} = e^{-\ln|x|} = e^{\ln|x|^{-1}} = |x|^{-1}$
Since $x \ge 1$, $|x| = x$, so $IF = \frac{1}{x}$.
Multiply the equation by the IF:
$\frac{1}{x}f'(x) - \frac{1}{x^2}f(x) = -1$
The left side is the derivative of the product of the IF and $f(x)$:
$\frac{d}{dx} \left( \frac{1}{x}f(x) \right) = -1$
Integrate both sides:
$\int \frac{d}{dx} \left( \frac{1}{x}f(x) \right) dx = \int -1 dx$
$\frac{1}{x}f(x) = -x + C$
where $C$ is the constant of integration.
The general solution is:
$f(x) = x(-x + C) = -x^2 + Cx$
Determine Constant C
Use the original equation with $x=1$ to find a condition for $f(x)$.
$6\int_1^1 f(t)dt = 3(1)f(1) + 1^3 - 4$
$0 = 3f(1) + 1 - 4$
$0 = 3f(1) - 3 \implies f(1) = 1$
Now use the general solution $f(x) = -x^2 + Cx$ with $f(1) = 1$:
$1 = -(1)^2 + C(1) \implies 1 = -1 + C \implies C = 2$
The specific solution is $f(x) = -x^2 + 2x$.
Calculate Value $f(2) - f(3)$
Evaluate the function at $x=2$ and $x=3$:
$f(2) = -(2)^2 + 2(2) = -4 + 4 = 0$
$f(3) = -(3)^2 + 2(3) = -9 + 6 = -3$
Calculate the difference:
$f(2) - f(3) = 0 - (-3) = 3$
The calculation yields $f(2) - f(3) = 3$.
Note: The provided correct answer is $-4$.
Similar Questions
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Important Questions from Integral Calculus
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