Questions: The equation is: (1/2)f''+ 4 (1/t)f'+ (k/t^2)f = 200, where t is the time in minutes, and k is a positive constant. (e) Let f(t) = t^n, where n is a constant. Find the value of n.

Solution Steps

Given the function \( f(t) = t^n \), we need to find its first and second derivatives:

\[ f'(t) = \frac{d}{dt}(t^n) = n t^{n-1} \]

\[ f''(t) = \frac{d^2}{dt^2}(t^n) = n(n-1) t^{n-2} \]

Substitute these into the given differential equation:

\[ \frac{1}{2}f'' + 4 \frac{1}{t}f' + \frac{k}{t^2}f = 200 \]

\[ \frac{1}{2} \cdot n(n-1) t^{n-2} + 4 \cdot \frac{1}{t} \cdot n t^{n-1} + \frac{k}{t^2} \cdot t^n = 200 \]

Simplify each term:

1. \(\frac{1}{2} n(n-1) t^{n-2}\) remains as is.
2. \(4 \cdot \frac{1}{t} \cdot n t^{n-1} = 4n t^{n-2}\).
3. \(\frac{k}{t^2} \cdot t^n = k t^{n-2}\).

Combine these terms:

\[ \frac{1}{2} n(n-1) t^{n-2} + 4n t^{n-2} + k t^{n-2} = 200 \]

Factor out \( t^{n-2} \):

\[ \left(\frac{1}{2} n(n-1) + 4n + k\right) t^{n-2} = 200 \]

For the equation to hold for all \( t \), the power of \( t \) on both sides must match. Since the right side is a constant (200), the power of \( t \) on the left must be zero:

\[ n-2 = 0 \implies n = 2 \]

Substitute \( n = 2 \) back into the coefficient equation:

\[ \frac{1}{2} \cdot 2(2-1) + 4 \cdot 2 + k = 200 \]

\[ 1 + 8 + k = 200 \]

\[ k = 200 - 9 = 191 \]

Final Answer