A particular solution of ` log ( dy //dx) = 3x + 5y, y(0) =0` is :

Options

A. `e^(x) + 3e^(-4y )=4`
B. `4e ^(3x) - 3e ^(-4y) = 3`
C. ` 3e^(3x)+4e^(4y)=7`
D. none of these

Answer

To solve the differential equation given by \( \log\left(\frac{dy}{dx}\right) = 3x + 5y \) with the initial condition \( y(0) = 0 \), we can follow these steps:

### Step 1: Remove the logarithm
We start by rewriting the equation in exponential form. This gives us:
\[ \frac{dy}{dx} = e^{3x + 5y} \]

### Step 2: Separate variables
Next, we separate the variables \( y \) and \( x \):
\[ \frac{dy}{e^{5y}} = e^{3x} \, dx \]

### Step 3: Rewrite the left side
We can rewrite the left side to make integration easier:
\[ e^{-5y} \, dy = e^{3x} \, dx \]

### Step 4: Integrate both sides
Now we integrate both sides:
\[ \int e^{-5y} \, dy = \int e^{3x} \, dx \]
The integration results in:
\[ -\frac{1}{5} e^{-5y} = \frac{1}{3} e^{3x} + C \]

### Step 5: Solve for \( C \) using the initial condition
We use the initial condition \( y(0) = 0 \) to find \( C \):
\[ -\frac{1}{5} e^{-5(0)} = \frac{1}{3} e^{3(0)} + C \]
This simplifies to:
\[ -\frac{1}{5} = \frac{1}{3} + C \]
Solving for \( C \):
\[ C = -\frac{1}{5} - \frac{1}{3} = -\frac{3}{15} - \frac{5}{15} = -\frac{8}{15} \]

### Step 6: Substitute \( C \) back into the equation
Now we substitute \( C \) back into our equation:
\[ -\frac{1}{5} e^{-5y} = \frac{1}{3} e^{3x} - \frac{8}{15} \]

### Step 7: Multiply through by -5
To simplify, we multiply through by -5:
\[ e^{-5y} = -\frac{5}{3} e^{3x} + \frac{8}{3} \]

### Step 8: Solve for \( y \)
Taking the natural logarithm of both sides to solve for \( y \):
\[ -5y = \log\left(-\frac{5}{3} e^{3x} + \frac{8}{3}\right) \]
Thus,
\[ y = -\frac{1}{5} \log\left(-\frac{5}{3} e^{3x} + \frac{8}{3}\right) \]

### Final Answer
The particular solution of the differential equation is:
\[ y = -\frac{1}{5} \log\left(-\frac{5}{3} e^{3x} + \frac{8}{3}\right) \]

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