Questions: Question 5 (1 point) Which of the following is a solution of the initial value problem y ln(y) dx + x dy = 0 y(e) = e Hint: ∫ 1/(t ln(t)) dt = ln(ln(t)) + K y = x e^(x-e) y = e^(x / e) y = e^(e / x) y = e^(x^2 / e^2) y = x e^(e-x)

Solution Steps

To solve the initial value problem, we need to separate the variables and integrate both sides. We will use the given hint to help with the integration. After integrating, we will apply the initial condition \( y(e) = e \) to find the constant of integration. Finally, we will check which of the given options satisfies the resulting equation.

1. Separate the variables in the differential equation.
2. Integrate both sides using the given hint.
3. Apply the initial condition to find the constant of integration.
4. Check which of the given options satisfies the resulting equation.

We start with the initial value problem given by the equation:

\[ y \ln(y) \, dx + x \, dy = 0 \]

We can rearrange this to separate the variables:

\[ \ln(\ln(y)) = -\ln(x) + C \]

Integrating both sides leads us to:

\[ \ln(x) + \ln(\ln(y)) = C \]

Using the initial condition \( y(e) = e \), we substitute \( x = e \) and \( y = e \):

\[ \ln(e) + \ln(\ln(e)) = 1 + 0 = 1 \]

Thus, we find that \( C = 1 \).

Substituting \( C \) back into the integrated equation gives us:

\[ \ln(x) + \ln(\ln(y)) + 1 = 0 \]

This simplifies to:

\[ \ln(\ln(y)) = -\ln(x) - 1 \]

We need to check which of the following options satisfies the final equation:

1. \( y = x e^{x - e} \)
2. \( y = e^{x/e} \)
3. \( y = e^{E/x} \)
4. \( y = e^{x^2/e^2} \)
5. \( y = x e^{e - x} \)

After evaluating, we find that the only option that satisfies the equation is:

\[ y = x e^{e - x} \]

Final Answer