`log ((dy)/(dx)) =2x +3y `

Answer

To solve the differential equation given by

\[ \log\left(\frac{dy}{dx}\right) = 2x + 3y, \]

we will follow these steps:

### Step 1: Exponentiate both sides
We start by exponentiating both sides to eliminate the logarithm:

\[ \frac{dy}{dx} = e^{2x + 3y}. \]

### Step 2: Rearranging the equation
Next, we can rearrange the equation to separate the variables \(y\) and \(x\):

\[ \frac{dy}{e^{3y}} = e^{2x} dx. \]

### Step 3: Integrate both sides
Now, we integrate both sides. The left side requires integration with respect to \(y\) and the right side with respect to \(x\):

\[ \int \frac{dy}{e^{3y}} = \int e^{2x} dx. \]

The left side can be rewritten as:

\[ \int e^{-3y} dy = \int e^{2x} dx. \]

### Step 4: Solve the integrals
Now we compute the integrals:

- For the left side:
\[ \int e^{-3y} dy = -\frac{1}{3} e^{-3y} + C_1, \]

- For the right side:
\[ \int e^{2x} dx = \frac{1}{2} e^{2x} + C_2. \]

### Step 5: Combine the results
Setting the results of the integrals equal to each other gives us:

\[ -\frac{1}{3} e^{-3y} = \frac{1}{2} e^{2x} + C, \]

where \(C = C_2 - C_1\) is a constant.

### Step 6: Rearranging the equation
To express \(e^{-3y}\) in terms of \(x\), we can multiply through by -3:

\[ e^{-3y} = -\frac{3}{2} e^{2x} - 3C. \]

### Step 7: Solve for \(y\)
Taking the natural logarithm of both sides to isolate \(y\):

\[ -3y = \log\left(-\frac{3}{2} e^{2x} - 3C\right), \]

thus,

\[ y = -\frac{1}{3} \log\left(-\frac{3}{2} e^{2x} - 3C\right). \]

### Final Solution
The final solution to the differential equation is:

\[ y = -\frac{1}{3} \log\left(-\frac{3}{2} e^{2x} - 3C\right). \]

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