what is the solution of the differential equation`ln ((dy)/(dx)) =ax +by` ?

Options

A. `a" "e^(ax)+1/be^(by)=c`
B. `1/ae^(ax)+1/be^(by)=c`
C. `a" "e^(ax)+1/be^(-by)=c`
D. `1/ae^(ax)+1/be^(-by)=c`

Explanation

To solve the differential equation given by

\[ \ln\left(\frac{dy}{dx}\right) = ax + by, \]

we will follow these steps:

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

\[ \frac{dy}{dx} = e^{ax + by}. \]

### Step 2: Rewrite the equation
We can rewrite the equation as:

\[ \frac{dy}{dx} = e^{ax} \cdot e^{by}. \]

### Step 3: Separate variables
Next, we separate the variables \(y\) and \(x\):

\[ \frac{dy}{e^{by}} = e^{ax} \, dx. \]

### Step 4: Integrate both sides
Now we integrate both sides. The left side requires the integral of \(e^{-by}\) and the right side requires the integral of \(e^{ax}\):

\[ \int e^{-by} \, dy = \int e^{ax} \, dx. \]

The left side integrates to:

\[ -\frac{1}{b} e^{-by} + C_1, \]

and the right side integrates to:

\[ \frac{1}{a} e^{ax} + C_2. \]

### Step 5: Combine the constants
We can combine the constants of integration into a single constant \(C\):

\[ -\frac{1}{b} e^{-by} = \frac{1}{a} e^{ax} + C. \]

### Step 6: Solve for \(y\)
To express \(y\) explicitly, we can rearrange the equation:

\[ e^{-by} = -\frac{b}{a} e^{ax} - bC. \]

Taking the natural logarithm of both sides gives us:

\[ -by = \ln\left(-\frac{b}{a} e^{ax} - bC\right). \]

Finally, we solve for \(y\):

\[ y = -\frac{1}{b} \ln\left(-\frac{b}{a} e^{ax} - bC\right). \]

This is the general solution of the differential equation.