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.
Transcript text: The equation is: $\frac{1}{2}f''+ 4 \frac{1}{t}f'+ \frac{k}{t^2}f = 200$, 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
Solution Steps
Step 1: Substitute f(t)=tn into the equation
Given the function f(t)=tn, we need to find its first and second derivatives:
f′(t)=dtd(tn)=ntn−1
f′′(t)=dt2d2(tn)=n(n−1)tn−2
Substitute these into the given differential equation:
21f′′+4t1f′+t2kf=200
21⋅n(n−1)tn−2+4⋅t1⋅ntn−1+t2k⋅tn=200
Step 2: Simplify the equation
Simplify each term:
21n(n−1)tn−2 remains as is.
4⋅t1⋅ntn−1=4ntn−2.
t2k⋅tn=ktn−2.
Combine these terms:
21n(n−1)tn−2+4ntn−2+ktn−2=200
Factor out tn−2:
(21n(n−1)+4n+k)tn−2=200
Step 3: Solve for n
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⟹n=2
Substitute n=2 back into the coefficient equation: