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
- Let a differentiable function $f$ satisfy the equation $\int_{0}^{36} f\left(\frac{tx}{36}\right) dt = 4\alpha f(x)$. If $y = f(x)$ is a standard parabola passing through the points (2, 1) and (– 4, $\beta$), then $\beta^\alpha$ is equal to ______.
- Let $[\cdot]$ denote the greatest integer function and $f(x) = \lim_{n \to \infty} \frac{1}{n^3} \sum_{k=1}^n \left[ \frac{k^2}{3^x} \right]$. Then $12 \sum_{j=1}^\infty f(j)$ is equal to __________.
Let $f$ be a differentiable function satisfying $f(x) = 1 - 2x + \int_{0}^{x} e^{(x - t)} f(t) dt$, $x \in \mathbf{R}$ and let $g(x) = \int_{0}^{x} (f(t) + 2)^{15} (t - 4)^{6} (t + 12)^{17} dt$, $x \in \mathbf{R}$. If $\text{p}$ and $\text{q}$ are respectively the points of local minima and local maxima of $g$, then the value of $|\text{p} + \text{q}|$ is equal to _________.
Let $f$ be a twice differentiable function such that $f(x) = \int_{0}^{x} \tan(t - x)dt - \int_{0}^{x} f(t)\tan t dt$, $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
Then $f''\left(\frac{\pi}{6}\right) + 12f'\left(-\frac{\pi}{6}\right) + f\left(\frac{\pi}{6}\right)$ is equal to __________- Let $y = y(x)$ be the solution of the differential equation $\left(x^2 - x\sqrt{x^2 - 1}\right)dy + \left(y\left(x - \sqrt{x^2 - 1}\right) - x\right)dx = 0, x \geq 1$. If $y(1) = 1$, then the greatest integer less than $y\left(\sqrt{5}\right)$ is _________.
- If $I_1 = \int_0^1 e^{-x} cos^2x dx$, $I_2 = \int_0^1 e^{-x^2} cos^2x dx$ and $I_3 = \int_0^1 e^{-x^2} dx$; then :
- If $\int \frac{(\sqrt{1+x^2}+x)^{10}}{(\sqrt{1+x^2}-x)^9} dx = \frac{1}{m} \left( (\sqrt{1+x^2}+x)^n (n\sqrt{1+x^2}-x) \right) + C$ where $C$ is the constant of integration and $m, n \in N$, then $m+n$ is equal to
Let [.] denote the greatest integer function. If $\int_{0}^{e^3} \left[\frac{1}{e^{x-1}}\right] dx = \alpha - \log_e 2$, then $\alpha^3$ is equal to ____________.
Let $f: R\to R$ be a thrice differentiable odd function satisfying $f'(x)\ge0, f''(x)=f(x), f(0)=0, f'(0)=3$. Then $9f(\log_e 3)$ is equal to ___________.
- Let $y = y (x)$ be the solution curve of the differentialequation $x (x^2 + e^x) dy + (e^x (x-2) y-x^3) dx = 0, x > 0$, passing through the point $(1, 0)$.Then $y (2)$ is equal to
Important Questions from Integral Calculus
- Let a differentiable function $f$ satisfy the equation $\int_{0}^{36} f\left(\frac{tx}{36}\right) dt = 4\alpha f(x)$. If $y = f(x)$ is a standard parabola passing through the points (2, 1) and (– 4, $\beta$), then $\beta^\alpha$ is equal to ______.
- Let $[\cdot]$ denote the greatest integer function and $f(x) = \lim_{n \to \infty} \frac{1}{n^3} \sum_{k=1}^n \left[ \frac{k^2}{3^x} \right]$. Then $12 \sum_{j=1}^\infty f(j)$ is equal to __________.
Let $f$ be a differentiable function satisfying $f(x) = 1 - 2x + \int_{0}^{x} e^{(x - t)} f(t) dt$, $x \in \mathbf{R}$ and let $g(x) = \int_{0}^{x} (f(t) + 2)^{15} (t - 4)^{6} (t + 12)^{17} dt$, $x \in \mathbf{R}$. If $\text{p}$ and $\text{q}$ are respectively the points of local minima and local maxima of $g$, then the value of $|\text{p} + \text{q}|$ is equal to _________.
Let $f$ be a twice differentiable function such that $f(x) = \int_{0}^{x} \tan(t - x)dt - \int_{0}^{x} f(t)\tan t dt$, $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
Then $f''\left(\frac{\pi}{6}\right) + 12f'\left(-\frac{\pi}{6}\right) + f\left(\frac{\pi}{6}\right)$ is equal to __________- Let $y = y(x)$ be the solution of the differential equation $\left(x^2 - x\sqrt{x^2 - 1}\right)dy + \left(y\left(x - \sqrt{x^2 - 1}\right) - x\right)dx = 0, x \geq 1$. If $y(1) = 1$, then the greatest integer less than $y\left(\sqrt{5}\right)$ is _________.
