Key Concepts and Formulas

• First-Order Linear Differential Equation: A differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$, where $P(y)$ and $Q(y)$ are functions of $y$.

• Integrating Factor: For a first-order linear differential equation in the form above, the integrating factor is given by $e^{\int P(y) dy}$.

• General Solution: The general solution is given by $x \cdot (\text{Integrating Factor}) = \int Q(y) \cdot (\text{Integrating Factor}) \, dy + C$, where $C$ is the constant of integration.

Step-by-Step Solution

Step 1: Rewrite the differential equation in the standard form

We are given the differential equation: $2(y+2) \log _{e}(y+2) d x+\left(x+4-2 \log _{e}(y+2)\right) d y=0$ Divide the entire equation by $dy$ to get: $2(y+2) \log _{e}(y+2) \frac{dx}{dy} + x + 4 - 2\log_e(y+2) = 0$ Rearrange the terms to isolate $\frac{dx}{dy}$: $2(y+2) \log _{e}(y+2) \frac{dx}{dy} + x = 2\log_e(y+2) - 4$ Now, divide by $2(y+2) \log _{e}(y+2)$ to get the standard form: $\frac{dx}{dy} + \frac{1}{2(y+2) \log _{e}(y+2)}x = \frac{2\log_e(y+2) - 4}{2(y+2) \log _{e}(y+2)}$ $\frac{dx}{dy} + \frac{1}{2(y+2) \log _{e}(y+2)}x = \frac{1}{y+2} - \frac{2}{(y+2)\log_e(y+2)}$

Step 2: Identify $P(y)$ and $Q(y)$

From the standard form, we have: $P(y) = \frac{1}{2(y+2) \log _{e}(y+2)}$ $Q(y) = \frac{1}{y+2} - \frac{2}{(y+2)\log_e(y+2)}$

Step 3: Calculate the Integrating Factor

The integrating factor is given by $e^{\int P(y) dy}$: $\int P(y) dy = \int \frac{1}{2(y+2) \log _{e}(y+2)} dy$ Let $u = \log_e(y+2)$. Then $du = \frac{1}{y+2} dy$. So, $\int P(y) dy = \int \frac{1}{2u} du = \frac{1}{2} \log_e |u| = \frac{1}{2} \log_e |\log_e(y+2)|$ The integrating factor is: $e^{\int P(y) dy} = e^{\frac{1}{2} \log_e (\log_e(y+2))} = \left( \log_e(y+2) \right)^{\frac{1}{2}} = \sqrt{\log_e(y+2)}$

Step 4: Find the general solution

The general solution is given by: $x \cdot (\text{Integrating Factor}) = \int Q(y) \cdot (\text{Integrating Factor}) \, dy + C$ $x \sqrt{\log_e(y+2)} = \int \left(\frac{1}{y+2} - \frac{2}{(y+2)\log_e(y+2)}\right) \sqrt{\log_e(y+2)} \, dy + C$ $x \sqrt{\log_e(y+2)} = \int \frac{\sqrt{\log_e(y+2)}}{y+2} dy - \int \frac{2\sqrt{\log_e(y+2)}}{(y+2)\log_e(y+2)} dy + C$ $x \sqrt{\log_e(y+2)} = \int \frac{\sqrt{\log_e(y+2)}}{y+2} dy - \int \frac{2}{(y+2)\sqrt{\log_e(y+2)}} dy + C$ Let $u = \log_e(y+2)$. Then $du = \frac{1}{y+2} dy$. $x \sqrt{\log_e(y+2)} = \int \sqrt{u} \, du - \int \frac{2}{\sqrt{u}} \, du + C$ $x \sqrt{\log_e(y+2)} = \frac{2}{3} u^{\frac{3}{2}} - 4\sqrt{u} + C$ $x \sqrt{\log_e(y+2)} = \frac{2}{3} (\log_e(y+2))^{\frac{3}{2}} - 4\sqrt{\log_e(y+2)} + C$ Divide by $\sqrt{\log_e(y+2)}$: $x = \frac{2}{3} \log_e(y+2) - 4 + \frac{C}{\sqrt{\log_e(y+2)}}$

Step 5: Apply the initial condition $x(e^4 - 2) = 1$

We are given that when $y = e^4 - 2$, $x = 1$. Substituting these values into the general solution: $1 = \frac{2}{3} \log_e(e^4 - 2 + 2) - 4 + \frac{C}{\sqrt{\log_e(e^4 - 2 + 2)}}$ $1 = \frac{2}{3} \log_e(e^4) - 4 + \frac{C}{\sqrt{\log_e(e^4)}}$ $1 = \frac{2}{3}(4) - 4 + \frac{C}{\sqrt{4}}$ $1 = \frac{8}{3} - 4 + \frac{C}{2}$ $1 = \frac{8 - 12}{3} + \frac{C}{2}$ $1 = -\frac{4}{3} + \frac{C}{2}$ $\frac{C}{2} = 1 + \frac{4}{3} = \frac{7}{3}$ $C = \frac{14}{3}$

Step 6: Find the particular solution

Substitute the value of $C$ back into the general solution: $x = \frac{2}{3} \log_e(y+2) - 4 + \frac{14}{3\sqrt{\log_e(y+2)}}$

Step 7: Evaluate $x(e^9 - 2)$

We need to find the value of $x$ when $y = e^9 - 2$: $x(e^9 - 2) = \frac{2}{3} \log_e(e^9 - 2 + 2) - 4 + \frac{14}{3\sqrt{\log_e(e^9 - 2 + 2)}}$ $x(e^9 - 2) = \frac{2}{3} \log_e(e^9) - 4 + \frac{14}{3\sqrt{\log_e(e^9)}}$ $x(e^9 - 2) = \frac{2}{3}(9) - 4 + \frac{14}{3\sqrt{9}}$ $x(e^9 - 2) = 6 - 4 + \frac{14}{3(3)}$ $x(e^9 - 2) = 2 + \frac{14}{9}$ $x(e^9 - 2) = \frac{18 + 14}{9} = \frac{32}{9}$

Step 8: Check for Errors and Correct There must be an error. Let's review the problem from the beginning. $2(y+2) \log _{e}(y+2) d x+\left(x+4-2 \log _{e}(y+2)\right) d y=0$ $2(y+2) \log _{e}(y+2) \frac{dx}{dy} + x + 4 - 2\log_e(y+2) = 0$ $2(y+2) \log _{e}(y+2) \frac{dx}{dy} + x = 2\log_e(y+2) - 4$ $\frac{dx}{dy} + \frac{1}{2(y+2) \log _{e}(y+2)}x = \frac{1}{y+2} - \frac{2}{(y+2)\log_e(y+2)}$ $P(y) = \frac{1}{2(y+2) \log _{e}(y+2)}$ $Q(y) = \frac{1}{y+2} - \frac{2}{(y+2)\log_e(y+2)}$ Integrating factor: $\sqrt{\log_e(y+2)}$ $x \sqrt{\log_e(y+2)} = \int \left( \frac{1}{y+2} - \frac{2}{(y+2)\log_e(y+2)}\right) \sqrt{\log_e(y+2)} dy$ $x \sqrt{\log_e(y+2)} = \int \frac{\sqrt{\log_e(y+2)}}{y+2} dy - \int \frac{2}{\sqrt{\log_e(y+2)}(y+2)} dy$ Let $u = \log_e(y+2)$, $du = \frac{dy}{y+2}$. $x \sqrt{u} = \int \sqrt{u} du - \int \frac{2}{\sqrt{u}} du$ $x \sqrt{u} = \frac{2}{3} u^{3/2} - 4\sqrt{u} + C$ $x = \frac{2}{3} u - 4 + C u^{-1/2}$ $x = \frac{2}{3} \log_e(y+2) - 4 + \frac{C}{\sqrt{\log_e(y+2)}}$ Now, $x(e^4 - 2) = 1$. $1 = \frac{2}{3} \log_e(e^4 - 2 + 2) - 4 + \frac{C}{\sqrt{\log_e(e^4 - 2 + 2)}}$ $1 = \frac{2}{3}(4) - 4 + \frac{C}{\sqrt{4}}$ $1 = \frac{8}{3} - 4 + \frac{C}{2}$ $\frac{7}{3} = \frac{C}{2}$ $C = \frac{14}{3}$ So, $x = \frac{2}{3} \log_e(y+2) - 4 + \frac{14}{3\sqrt{\log_e(y+2)}}$ Now, we evaluate $x(e^9 - 2)$. $x(e^9 - 2) = \frac{2}{3} \log_e(e^9) - 4 + \frac{14}{3\sqrt{\log_e(e^9)}}$ $x(e^9 - 2) = \frac{2}{3}(9) - 4 + \frac{14}{3\sqrt{9}}$ $x(e^9 - 2) = 6 - 4 + \frac{14}{9} = 2 + \frac{14}{9} = \frac{32}{9}$ There seems to be no error in the calculation, which contradicts the correct answer.

Step 9: Re-Examine the Solution Let's try to see if $x(e^9 - 2) = \frac{4}{9}$ is possible, and if so, what would the constant $C$ have to be. $\frac{4}{9} = \frac{2}{3}(9) - 4 + \frac{C}{3}$ $\frac{4}{9} = 6 - 4 + \frac{C}{3}$ $\frac{4}{9} = 2 + \frac{C}{3}$ $-\frac{14}{9} = \frac{C}{3}$ $C = -\frac{14}{3}$ If this is true, then $1 = \frac{2}{3} (4) - 4 + \frac{C}{2}$ $1 = \frac{8}{3} - 4 + \frac{C}{2}$ $\frac{7}{3} = \frac{C}{2}$ $C = \frac{14}{3}$ There seems to be an error in the answer key.

Step 10: Final Solution Since the derivation is correct, and the answer key seems wrong, let's proceed with the calculated value.

Common Mistakes & Tips

• Integration Errors: Be extremely careful while performing the integration, especially when using substitution. Double-check your substitutions and the resulting integrals.
• Algebraic Errors: Pay close attention to signs and coefficients when simplifying expressions and solving for the constant of integration. A small error can propagate through the entire solution.
• Checking the Solution: After finding the particular solution, it's a good practice to plug it back into the original differential equation to verify that it satisfies the equation and the initial condition.

Summary

We solved the given first-order linear differential equation by first rewriting it in the standard form, then finding the integrating factor, and finally obtaining the general solution. We then applied the initial condition to determine the constant of integration and find the particular solution. Finally, we evaluated the particular solution at $y = e^9 - 2$. The correct answer appears to be $\frac{32}{9}$.

Final Answer

The final answer is \boxed{\frac{32}{9}}, which corresponds to option (B).