Related Formula
Leibniz's Rule: (d)/(dx) ∫ₐx f(t) dt = f(x)$$\text{Leibniz's Rule: } \frac{d}{dx} \int_{a}^{x} f(t) dt = f(x)$$
Linear DE form: (dy)/(dx) + P(x)y = Q(x)$$\text{Linear DE form: } \frac{dy}{dx} + P(x)y = Q(x)$$
Core Logic
Given 6∫₁xf(t)dt=3xf(x)+x³-4$6\int_{1}^{x}f(t)dt=3xf(x)+x^{3}-4$.
Differentiating both sides with respect to x$x$:
6 f(x) = 3(1 · f(x) + x f'(x)) + 3x²$$6 f(x) = 3(1 \cdot f(x) + x f'(x)) + 3x^2$$
6 f(x) = 3f(x) + 3x f'(x) + 3x²$$6 f(x) = 3f(x) + 3x f'(x) + 3x^2$$
3f(x) = 3x f'(x) + 3x²$$3f(x) = 3x f'(x) + 3x^2$$
Step 1: Solving the Differential Equation
Let y = f(x)$y = f(x)$. The equation simplifies to:
x (dy)/(dx) - y = -x²$$x \frac{dy}{dx} - y = -x^2$$
Divide the entire equation by x²$x^2$ to recognize the exact differential form:
(x (dy)/(dx) - y)/(x²) = -1$$\frac{x \frac{dy}{dx} - y}{x^2} = -1$$
(d)/(dx) ( (y)/(x) ) = -1$$\frac{d}{dx} \left( \frac{y}{x} \right) = -1$$
Integrate both sides:
(y)/(x) = -x + C$$\frac{y}{x} = -x + C$$
y = -x² + Cx f(x) = -x² + Cx$$y = -x^2 + Cx \implies f(x) = -x^2 + Cx$$
Step 2: Evaluating the Constant
To find C$C$, we plug x=1$x=1$ into the original integral equation:
6∫₁¹f(t)dt = 3(1)f(1) + 1³ - 4$$6\int_{1}^{1}f(t)dt = 3(1)f(1) + 1^3 - 4$$
0 = 3f(1) - 3 f(1) = 1$$0 = 3f(1) - 3 \implies f(1) = 1$$
Using f(x) = -x² + Cx$f(x) = -x^2 + Cx$:
f(1) = -1 + C = 1 C = 2$$f(1) = -1 + C = 1 \implies C = 2$$
Thus, f(x) = -x² + 2x$f(x) = -x^2 + 2x$.
Step 3: Final Calculation
We need to find f(2) - f(3)$f(2) - f(3)$:
f(2) = -(2)² + 2(2) = -4 + 4 = 0$$f(2) = -(2)^2 + 2(2) = -4 + 4 = 0$$
f(3) = -(3)² + 2(3) = -9 + 6 = -3$$f(3) = -(3)^2 + 2(3) = -9 + 6 = -3$$
f(2) - f(3) = 0 - (-3) = 3$$f(2) - f(3) = 0 - (-3) = 3$$
Pattern Recognition
An integral equation involving a generic function f(t)$f(t)$ within variable limits signals the immediate use of Leibniz's rule. Differentiating leads cleanly to a linear differential equation. Re-substituting the lower limit (x=1$x=1$) securely yields the boundary condition f(1)$f(1)$.
Chapter Mix
Class 12 Maths: Differential Equations
Class 12 Maths: Applications of Integrals
