A. A one-pound pack of wall putty costs $3. A 10-pound pack costs $24. What is the percent savings If you buy the 10-pound pack?
a. 20%
b. 24%
c. 25%
d. 60%
B. Solve the differential equation.
x2y” + 5xy’ + 4y = 0
The correct answer and explanation is:
Part A: Percent Savings
To calculate the percent savings when buying the 10-pound pack compared to the cost of buying 10 individual 1-pound packs:
- Cost of 10 individual 1-pound packs:
[
10 \times 3 = 30 \, \text{dollars}.
] - Cost of the 10-pound pack:
[
24 \, \text{dollars}.
] - Savings:
[
30 – 24 = 6 \, \text{dollars}.
] - Percent savings:
[
\left( \frac{\text{Savings}}{\text{Cost of 10 individual packs}} \right) \times 100 = \left( \frac{6}{30} \right) \times 100 = 20\%.
]
Correct answer:
a. 20%
Part B: Differential Equation Solution
We solve the given second-order linear differential equation:
[
x^2 y” + 5x y’ + 4y = 0.
]
Step 1: Identify the type of equation
This is a Cauchy-Euler equation because the coefficients of (y”), (y’), and (y) are powers of (x).
Step 2: Substitute (y = x^r)
Assume a solution of the form (y = x^r), where (r) is a constant to be determined.
Calculate derivatives:
[
y’ = r x^{r-1}, \quad y” = r(r-1)x^{r-2}.
]
Substitute these into the differential equation:
[
x^2 \cdot r(r-1)x^{r-2} + 5x \cdot r x^{r-1} + 4x^r = 0.
]
Simplify:
[
r(r-1)x^r + 5r x^r + 4x^r = 0.
]
Factor (x^r) (non-zero for (x > 0)):
[
r(r-1) + 5r + 4 = 0.
]
Step 3: Solve the characteristic equation
Expand and combine terms:
[
r^2 – r + 5r + 4 = 0 \quad \Rightarrow \quad r^2 + 4r + 4 = 0.
]
Factorize:
[
(r + 2)^2 = 0.
]
Thus, (r = -2) is a repeated root.
Step 4: General solution
For a repeated root (r = -2), the general solution is:
[
y(x) = C_1 x^{-2} + C_2 x^{-2} \ln(x),
]
where (C_1) and (C_2) are arbitrary constants.
Explanation
The given differential equation is a second-order linear Cauchy-Euler equation, characterized by variable coefficients proportional to powers of (x). These equations are efficiently solved using the substitution (y = x^r), which transforms the original equation into an algebraic equation called the characteristic equation.
In this problem, substituting (y = x^r) led to the characteristic equation (r^2 + 4r + 4 = 0), which is quadratic. Solving this yields (r = -2) as a repeated root. When the characteristic equation has repeated roots, the general solution incorporates a logarithmic term, giving (y(x) = C_1 x^{-2} + C_2 x^{-2} \ln(x)).
This solution represents the superposition of two independent solutions: one proportional to (x^{-2}) and another to (x^{-2} \ln(x)). These forms arise from the underlying properties of second-order differential equations and their solutions in terms of linearly independent functions.
The method and solution reflect the importance of matching the solution’s form to the equation’s structure. Cauchy-Euler equations frequently model physical systems with scale-invariant properties, such as heat conduction or wave propagation in certain media.