Key Concepts and Formulas

• First-Order Linear Differential Equation: A differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$.
• Integrating Factor (IF): For a first-order linear differential equation, the integrating factor is $IF = e^{\int P(x) dx}$.
• Solution using Integrating Factor: The solution to the differential equation is given by $y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$.

Step-by-Step Solution

Step 1: Rearranging the Differential Equation into Standard Form

We are given the differential equation: $2(x^2 + x^{5/4})dy - y(x + x^{1/4})dx = 2x^{9/4}dx$ Our goal is to rewrite this in the standard form $\frac{dy}{dx} + P(x)y = Q(x)$.

1. Divide by $dx$: $2(x^2 + x^{5/4})\frac{dy}{dx} - y(x + x^{1/4}) = 2x^{9/4}$

2. Divide by $2(x^2 + x^{5/4})$: $\frac{dy}{dx} - \frac{(x + x^{1/4})}{2(x^2 + x^{5/4})}y = \frac{2x^{9/4}}{2(x^2 + x^{5/4})}$

3. Simplify: $\frac{dy}{dx} - \frac{x + x^{1/4}}{2(x^2 + x^{5/4})}y = \frac{x^{9/4}}{x^2 + x^{5/4}}$

4. Identify $P(x)$ and $Q(x)$: $P(x) = -\frac{x + x^{1/4}}{2(x^2 + x^{5/4})}$ $Q(x) = \frac{x^{9/4}}{x^2 + x^{5/4}}$

Step 2: Simplifying $P(x)$ and $Q(x)$

Simplifying these expressions is crucial for manageable integration.

1. Simplify $P(x)$: $P(x) = -\frac{x^{1/4}(x^{3/4} + 1)}{2x^{5/4}(x^{3/4} + 1)} = -\frac{1}{2x}$

2. Simplify $Q(x)$: $Q(x) = \frac{x^{9/4}}{x^{5/4}(x^{3/4} + 1)} = \frac{x}{x^{3/4} + 1}$

Step 3: Calculating the Integrating Factor (IF)

1. Calculate $\int P(x) dx$: $\int P(x) dx = \int -\frac{1}{2x} dx = -\frac{1}{2} \int \frac{1}{x} dx = -\frac{1}{2} \ln(x)$

2. Calculate the Integrating Factor: $IF = e^{\int P(x) dx} = e^{-\frac{1}{2} \ln(x)} = e^{\ln(x^{-1/2})} = x^{-1/2} = \frac{1}{\sqrt{x}}$

Step 4: Finding the General Solution

1. Apply the formula $y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$: $y \cdot \frac{1}{\sqrt{x}} = \int \frac{x}{x^{3/4} + 1} \cdot \frac{1}{\sqrt{x}} dx + C$ $\frac{y}{\sqrt{x}} = \int \frac{\sqrt{x}}{x^{3/4} + 1} dx + C$

2. Solve the integral: Let $x^{1/4} = t$, so $x = t^4$ and $dx = 4t^3 dt$. $\int \frac{\sqrt{x}}{x^{3/4} + 1} dx = \int \frac{t^2}{t^3 + 1} \cdot 4t^3 dt = 4 \int \frac{t^5}{t^3 + 1} dt$

3. Perform polynomial long division: $\frac{t^5}{t^3+1} = t^2 - \frac{t^2}{t^3+1}$ so $\frac{t^5}{t^3+1} = t^2 - \frac{t^2}{t^3+1}$. $4 \int \frac{t^5}{t^3 + 1} dt = 4 \int \left(t^2 - \frac{t^2}{t^3 + 1}\right) dt = 4 \left(\int t^2 dt - \int \frac{t^2}{t^3 + 1} dt\right)$

4. Evaluate the integrals: $4 \left(\frac{t^3}{3} - \frac{1}{3} \ln|t^3 + 1|\right) + C_1 = \frac{4}{3}t^3 - \frac{4}{3} \ln(t^3 + 1) + C_1$ $= \frac{4}{3}x^{3/4} - \frac{4}{3} \ln(x^{3/4} + 1) + C_1$

5. Substitute back into the solution: $\frac{y}{\sqrt{x}} = \frac{4}{3}x^{3/4} - \frac{4}{3} \ln(x^{3/4} + 1) + C$ $y = \sqrt{x} \left(\frac{4}{3}x^{3/4} - \frac{4}{3} \ln(x^{3/4} + 1) + C\right)$ $y = \frac{4}{3}x^{5/4} - \frac{4}{3}\sqrt{x} \ln(x^{3/4} + 1) + C\sqrt{x}$

Step 5: Applying the Initial Condition

We are given that $y(1) = 1 - \frac{4}{3}\ln(2)$. Substitute $x = 1$ and $y = 1 - \frac{4}{3}\ln(2)$ into the equation:

$1 - \frac{4}{3}\ln(2) = \frac{4}{3}(1)^{5/4} - \frac{4}{3}\sqrt{1} \ln((1)^{3/4} + 1) + C\sqrt{1}$ $1 - \frac{4}{3}\ln(2) = \frac{4}{3} - \frac{4}{3}\ln(2) + C$ $C = 1 - \frac{4}{3} = -\frac{1}{3}$

Step 6: Finding the Particular Solution

Substitute $C = -\frac{1}{3}$ into the general solution: $y = \frac{4}{3}x^{5/4} - \frac{4}{3}\sqrt{x} \ln(x^{3/4} + 1) - \frac{1}{3}\sqrt{x}$

Step 7: Finding $y(16)$

Substitute $x = 16$ into the particular solution: $y(16) = \frac{4}{3}(16)^{5/4} - \frac{4}{3}\sqrt{16} \ln((16)^{3/4} + 1) - \frac{1}{3}\sqrt{16}$ $y(16) = \frac{4}{3}(2^4)^{5/4} - \frac{4}{3}(4) \ln((2^4)^{3/4} + 1) - \frac{1}{3}(4)$ $y(16) = \frac{4}{3}(2^5) - \frac{16}{3} \ln(2^3 + 1) - \frac{4}{3}$ $y(16) = \frac{4}{3}(32) - \frac{16}{3} \ln(9) - \frac{4}{3}$ $y(16) = \frac{128}{3} - \frac{16}{3} \ln(3^2) - \frac{4}{3}$ $y(16) = \frac{124}{3} - \frac{16}{3} (2\ln(3))$ $y(16) = \frac{124}{3} - \frac{32}{3} \ln(3)$ $y(16) = 4 \left(\frac{31}{3} - \frac{8}{3} \ln(3)\right)$

Common Mistakes & Tips

• Simplification is Key: Always simplify $P(x)$ and $Q(x)$ before integrating. This can significantly reduce the complexity of the integral.
• Substitution: When faced with a difficult integral, consider using a suitable substitution to simplify the expression.
• Careful Calculation: Pay close attention to arithmetic and algebraic manipulations to avoid errors.

Summary

We solved the given first-order linear differential equation by first rearranging it into the standard form, then finding the integrating factor. Using the integrating factor, we found the general solution and then used the given initial condition to determine the constant of integration, leading to the particular solution. Finally, we substituted $x = 16$ into the particular solution to find the value of $y(16)$.

The final answer is $4\left( {{{31} \over 3} - {8 \over 3}{{\log }_e}3} \right)$, which corresponds to option (A).

Final Answer

The final answer is $\boxed{4\left( {{{31} \over 3} - {8 \over 3}{{\log }_e}3} \right)}$, which corresponds to option (A).