`(dy)/(dx)=2e^(x)y^(3)`, when `x=0, y=(1)/(2)`

Options

A. `4e^(-x)+y^(2)=8`
B. `4e^(x)+y^(2)=8`
C. `4e^(x)+y^(-2)=8`
D. `4e^(y)-x^(-2)=8`

Answer

To solve the differential equation \(\frac{dy}{dx} = 2e^x y^3\) with the initial condition \(y(0) = \frac{1}{2}\), we will follow these steps:

### Step 1: Separate the Variables
We start by separating the variables \(y\) and \(x\):
\[ \frac{dy}{y^3} = 2e^x dx \]

### Step 2: Integrate Both Sides
Next, we integrate both sides. The left side integrates to:
\[ \int \frac{dy}{y^3} = \int 2e^x dx \]
The left side becomes:
\[ -\frac{1}{2y^2} \]
The right side integrates to:
\[ 2e^x + C \]
Thus, we have:
\[ -\frac{1}{2y^2} = 2e^x + C \]

### Step 3: Rearranging the Equation
We can rearrange the equation to isolate \(y^2\):
\[ \frac{1}{2y^2} = -2e^x - C \]
Multiplying through by -1 gives:
\[ \frac{1}{2y^2} = -2e^x + C \]

### Step 4: Solve for \(y^2\)
Now, we can solve for \(y^2\):
\[ y^2 = \frac{1}{2(-2e^x + C)} \]

### Step 5: Apply the Initial Condition
We will use the initial condition \(y(0) = \frac{1}{2}\) to find \(C\):
\[ \left(\frac{1}{2}\right)^2 = \frac{1}{2(-2e^0 + C)} \]
This simplifies to:
\[ \frac{1}{4} = \frac{1}{2(-2 + C)} \]
Cross-multiplying gives:
\[ 1 = 2(-2 + C) \]
Thus:
\[ 1 = -4 + 2C \implies 2C = 5 \implies C = \frac{5}{2} \]

### Step 6: Substitute \(C\) Back into the Equation
Now we substitute \(C\) back into the equation for \(y^2\):
\[ y^2 = \frac{1}{2(-2e^x + \frac{5}{2})} \]

### Step 7: Simplify the Equation
This simplifies to:
\[ y^2 = \frac{1}{-4e^x + 5} \]

### Step 8: Solve for \(y\)
Taking the square root gives:
\[ y = \frac{1}{\sqrt{-4e^x + 5}} \]

### Final Solution
Thus, the solution to the differential equation is:
\[ y = \frac{1}{\sqrt{-4e^x + 5}} \]