Question

Given that \[ \frac{dy}{dx}=ye^x \] and when \[ x=0,\quad y=e, \] then the value of \(y\) (\(y>0\)) when \(x=1\) is

• A. \[ \frac1e \]
• B. \[ e \]
• C. \[ e^e \]
• D. \[ \log e \] \bigskip

Hint:

For equations of the form \[ \frac{dy}{dx}=y\,f(x), \] separate variables: \[ \frac{dy}{y}=f(x)\,dx, \] integrate, and then apply the initial condition.

Answer 1

The Correct Option is C

Concept: The given differential equation is separable.

Step 1: Separate the variables. \[ \frac{dy}{dx}=ye^x. \] \[ \frac{dy}{y}=e^x\,dx. \]

Step 2: Integrate both sides. \[ \int \frac{dy}{y} = \int e^x\,dx. \] \[ \ln y = e^x+C. \] \[ y=e^{\,e^x+C} = Ke^{e^x}, \] where \[ K=e^C. \]

Step 3: Use the initial condition. Given \[ x=0,\qquad y=e. \] Substituting, \[ e = K e^{e^0} = K e. \] Hence \[ K=1. \] Therefore, \[ y=e^{e^x}. \]

Step 4: Find \(y\) when \(x=1\). \[ y=e^{e^1}. \] \[ y=e^e. \]

Step 5: Write the final answer. \[ \boxed{e^e} \]