A particular solution of log `((dy)/(dx))= 3x + 4 y , y (0) = 0 ` is

Options

A. `e^(3x) +3e^(-4y) =4`
B. ` 4e^(3x) - 3e^(-4y) =3`
C. ` 3e^(3x) +4e^(-4y) =7`
D. ` 4e^(3x)+3e^(-4y)=7`

Explanation

To find the particular solution of the differential equation given by \( \log\left(\frac{dy}{dx}\right) = 3x + 4y \) with the initial condition \( y(0) = 0 \), we can follow these steps:

### Step 1: Rewrite the equation
Starting from the given equation:
\[ \log\left(\frac{dy}{dx}\right) = 3x + 4y \]
we can exponentiate both sides to eliminate the logarithm:
\[ \frac{dy}{dx} = e^{3x + 4y} \]

### Step 2: Separate variables
We can rewrite the equation as:
\[ \frac{dy}{dx} = e^{3x} \cdot e^{4y} \]
Now, we can separate the variables \( y \) and \( x \):
\[ e^{-4y} dy = e^{3x} dx \]

### Step 3: Integrate both sides
Next, we integrate both sides:
\[ \int e^{-4y} dy = \int e^{3x} dx \]
The left side integrates to:
\[ -\frac{1}{4} e^{-4y} \]
And the right side integrates to:
\[ \frac{1}{3} e^{3x} + C \]
Thus, we have:
\[ -\frac{1}{4} e^{-4y} = \frac{1}{3} e^{3x} + C \]

### Step 4: Solve for \( y \)
Rearranging gives:
\[ e^{-4y} = -\frac{4}{3} e^{3x} - 4C \]
Taking the reciprocal:
\[ e^{4y} = \frac{1}{-\frac{4}{3} e^{3x} - 4C} \]
Taking the natural logarithm:
\[ 4y = \log\left(\frac{1}{-\frac{4}{3} e^{3x} - 4C}\right) \]
Thus:
\[ y = \frac{1}{4} \log\left(\frac{1}{-\frac{4}{3} e^{3x} - 4C}\right) \]

### Step 5: Apply the initial condition
Using the initial condition \( y(0) = 0 \):
\[ 0 = \frac{1}{4} \log\left(\frac{1}{-\frac{4}{3} e^{0} - 4C}\right) \]
This simplifies to:
\[ \log\left(\frac{1}{-\frac{4}{3} - 4C}\right) = 0 \]
Thus:
\[ -\frac{4}{3} - 4C = 1 \implies -4C = 1 + \frac{4}{3} = \frac{7}{3} \implies C = -\frac{7}{12} \]

### Step 6: Substitute \( C \) back into the equation
Substituting \( C \) back into our equation gives:
\[ y = \frac{1}{4} \log\left(\frac{1}{-\frac{4}{3} e^{3x} + \frac{7}{3}}\right) \]

### Final Result
Thus, the particular solution is:
\[ y = \frac{1}{4} \log\left(\frac{1}{-\frac{4}{3} e^{3x} + \frac{7}{3}}\right) \]