Question

If \(x \phi(x) = \int_{5}^{x} (3t^2 - 2\phi'(t))dt\), \(x>-2\), and \(\phi(0)=4\), then \(\phi(2)\) is ___________

Hint:

Notice that the left side \(\phi(x) + x\phi'(x)\) is almost the derivative of \(x\phi(x)\). Here, finding the I.F. leads directly to a very simple integration.

Answer 1

Correct Answer: 4

Step 1: Understanding the Concept:
We differentiate both sides with respect to \(x\) using the Newton-Leibniz formula to convert the integral equation into a linear differential equation.
Step 2: Detailed Explanation:
Differentiate both sides of \(x \phi(x) = \int_{5}^{x} (3t^2 - 2\phi'(t))dt\) w.r.t \(x\):
\[ \phi(x) + x \phi'(x) = 3x^2 - 2\phi'(x) \]
Rearrange the terms:
\[ (x + 2) \phi'(x) + \phi(x) = 3x^2 \]
This is a first-order linear differential equation. Divide by \((x+2)\):
\[ \frac{d\phi}{dx} + \frac{1}{x+2} \phi = \frac{3x^2}{x+2} \]
Integrating factor (I.F.) = \(e^{\int \frac{1}{x+2} dx} = e^{\ln(x+2)} = x+2\).
Multiplying the equation by I.F.:
\[ \frac{d}{dx} [\phi(x) \cdot (x+2)] = \frac{3x^2}{x+2} \cdot (x+2) = 3x^2 \]
Integrate both sides:
\[ \phi(x) \cdot (x+2) = \int 3x^2 dx = x^3 + C \]
Using initial condition \(\phi(0) = 4\):
\[ 4(0 + 2) = 0^3 + C \implies C = 8 \]
So, \(\phi(x) = \frac{x^3 + 8}{x + 2}\).
Using the identity \(x^3+8 = (x+2)(x^2-2x+4)\):
\[ \phi(x) = x^2 - 2x + 4 \]
Now, \(\phi(2) = 2^2 - 2(2) + 4 = 4 - 4 + 4 = 4\).
Step 3: Final Answer:
The value of \(\phi(2)\) is 4.