If y=f(x) is the solution of differential equation , `e^y((dy)/(dx)-2)=e^(3x)` such that f(0)=0 , then f(2) is equal to :

Options

A. 3
B. 5
C. 6
D. None of these

Explanation

To solve the differential equation \( e^y \left( \frac{dy}{dx} - 2 \right) = e^{3x} \) with the initial condition \( f(0) = 0 \), we will follow these steps:

### Step 1: Rewrite the differential equation
We start with the given equation:
\[ e^y \left( \frac{dy}{dx} - 2 \right) = e^{3x} \]
We can rearrange this to isolate \( \frac{dy}{dx} \):
\[ e^y \frac{dy}{dx} - 2e^y = e^{3x} \]
This gives us:
\[ e^y \frac{dy}{dx} = e^{3x} + 2e^y \]
Now, divide both sides by \( e^y \):
\[ \frac{dy}{dx} = e^{3x - y} + 2 \]

### Step 2: Introduce a substitution
Let \( t = e^y \). Then, \( \frac{dy}{dx} = \frac{dt}{dx} \cdot \frac{1}{t} \). Substituting this into the equation gives:
\[ \frac{1}{t} \frac{dt}{dx} = e^{3x - \ln t} + 2 \]
This simplifies to:
\[ \frac{1}{t} \frac{dt}{dx} = e^{3x} \cdot \frac{1}{t} + 2 \]
Multiplying through by \( t \) yields:
\[ \frac{dt}{dx} = e^{3x} + 2t \]

### Step 3: Rearranging into standard form
Rearranging gives us:
\[ \frac{dt}{dx} - 2t = e^{3x} \]
This is now in the standard form of a linear first-order differential equation.

### Step 4: Find the integrating factor
The integrating factor \( I \) is given by:
\[ I = e^{\int -2 \, dx} = e^{-2x} \]

### Step 5: Multiply through by the integrating factor
Multiplying the entire equation by the integrating factor:
\[ e^{-2x} \frac{dt}{dx} - 2 e^{-2x} t = e^{3x} e^{-2x} \]
This simplifies to:
\[ e^{-2x} \frac{dt}{dx} - 2 e^{-2x} t = e^{x} \]

### Step 6: Solve the left-hand side
The left-hand side can be expressed as:
\[ \frac{d}{dx} (e^{-2x} t) = e^{x} \]

### Step 7: Integrate both sides
Integrating both sides gives:
\[ e^{-2x} t = \int e^{x} \, dx = e^{x} + C \]
Thus,
\[ t = e^{2x} (e^{x} + C) = e^{3x} + Ce^{2x} \]

### Step 8: Substitute back for \( y \)
Recall that \( t = e^y \), so:
\[ e^y = e^{3x} + Ce^{2x} \]
Taking the natural logarithm:
\[ y = \ln(e^{3x} + Ce^{2x}) \]

### Step 9: Apply the initial condition
Using the initial condition \( f(0) = 0 \):
\[ 0 = \ln(e^{0} + C e^{0}) \implies 0 = \ln(1 + C) \implies 1 + C = 1 \implies C = 0 \]
Thus, we have:
\[ y = \ln(e^{3x}) = 3x \]

### Step 10: Find \( f(2) \)
Finally, substituting \( x = 2 \):
\[ f(2) = 3 \cdot 2 = 6 \]

### Final Answer
Thus, \( f(2) = 6 \).

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