Question

Let \(y : (-\infty, \infty) \to (0, \infty)\) be the solution of the differential equation \[ \frac{dy}{dx} = \frac{e^{5x}y^3 + y^3}{e^x + e^x y^4} \] satisfying \[ y(0) = \frac{1}{\sqrt{2}}. \] Then the value of \[ y(\log_e 2) \] is:

• A. $\sqrt{\frac{5 + \sqrt{35}}{2}}$
• B. $\sqrt{\frac{7 + \sqrt{53}}{2}}$
• C. $\frac{7 + \sqrt{53}}{2}$
• D. $\frac{5 + \sqrt{35}}{2}$

Hint:

Always check for symmetries or simplifications in the differential equation before integrating. Factoring out common terms like $y^3$ and $e^x$ here immediately reveals the separable structure.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The goal is to solve a first-order non-linear differential equation using the separation of variables method and then evaluate the function at a specific value of $x$.

Step 2: Key Formula or Approach:

• Rearrange the differential equation into the form $g(y)dy = f(x)dx$.

• Integrate both sides and use the initial condition to find the constant of integration.

Step 3: Detailed Explanation:

• Simplify the expression: \[ \frac{dy}{dx} = \frac{y^3(e^{5x} + 1)}{e^x(1 + y^4)} = (e^{4x} + e^{-x}) \frac{y^3}{1 + y^4} \]
• Separate variables: \[ \frac{1 + y^4}{y^3} dy = (e^{4x} + e^{-x}) dx \implies (y^{-3} + y) dy = (e^{4x} + e^{-x}) dx \]
• Integrate both sides: \[ \frac{y^{-2}}{-2} + \frac{y^2}{2} = \frac{e^{4x}}{4} - e^{-x} + C \implies \frac{y^2}{2} - \frac{1}{2y^2} = \frac{e^{4x}}{4} - e^{-x} + C \]
• Using initial condition $y(0) = 1/\sqrt{2} \implies y^2 = 1/2$: \[ \frac{1}{4} - \frac{1}{2(1/2)} = \frac{1}{4} - 1 + C \implies \frac{1}{4} - 1 = \frac{1}{4} - 1 + C \implies C = 0 \]
• Equation becomes: $y^2 - \frac{1}{y^2} = \frac{e^{4x}}{2} - 2e^{-x}$.

• At $x = \log_e 2$, $e^x = 2$, $e^{4x} = 16$, and $e^{-x} = 1/2$: \[ y^2 - \frac{1}{y^2} = \frac{16}{2} - 2(1/2) = 8 - 1 = 7 \]
• Let $t = y^2$: $t - \frac{1}{t} = 7 \implies t^2 - 7t - 1 = 0$.

• Solving for $t$ using the quadratic formula: $t = \frac{7 \pm \sqrt{49 - 4(1)(-1)}}{2} = \frac{7 \pm \sqrt{53}}{2}$.

• Since $y \in (0, \infty)$, $y^2 > 0$, we take the positive root: $y^2 = \frac{7 + \sqrt{53}}{2} \implies y = \sqrt{\frac{7 + \sqrt{53}}{2}}$.

Step 4: Final Answer:
The value of $y(\log_e 2)$ is $\sqrt{\frac{7 + \sqrt{53}}{2}}$.