To solve the differential equation given by
\[
\log\left(\frac{dy}{dx}\right) = ax + by,
\]
we will follow the steps outlined in the video transcript.
### Step 1: Exponentiate both sides
To eliminate the logarithm, we exponentiate both sides of the equation:
\[
\frac{dy}{dx} = e^{ax + by}.
\]
**Hint:** Remember that if \(\log(z) = k\), then \(z = e^k\).
### Step 2: Separate variables
Next, we will separate the variables \(y\) and \(x\). We can rewrite the equation as:
\[
dy = e^{ax + by} \, dx.
\]
Now, we need to isolate \(y\) terms on one side and \(x\) terms on the other. We can rewrite \(e^{ax + by}\) as:
\[
dy = e^{ax} e^{by} \, dx.
\]
Dividing both sides by \(e^{by}\):
\[
e^{-by} dy = e^{ax} \, dx.
\]
**Hint:** When separating variables, ensure that all \(y\) terms are on one side and all \(x\) terms on the other.
### Step 3: Integrate both sides
Now we integrate both sides:
\[
\int e^{-by} \, dy = \int e^{ax} \, dx.
\]
The left side integrates to:
\[
-\frac{1}{b} e^{-by} + C_1,
\]
and the right side integrates to:
\[
\frac{1}{a} e^{ax} + C_2.
\]
Combining the constants of integration, we can write:
\[
-\frac{1}{b} e^{-by} = \frac{1}{a} e^{ax} + C,
\]
where \(C = C_2 - C_1\).
**Hint:** When integrating, don't forget to include the constant of integration.
### Step 4: Solve for \(y\)
Now, we need to isolate \(y\). First, multiply through by \(-b\):
\[
e^{-by} = -\frac{b}{a} e^{ax} - bC.
\]
Taking the natural logarithm of both sides gives:
\[
-by = \log\left(-\frac{b}{a} e^{ax} - bC\right).
\]
Thus, we can express \(y\) as:
\[
y = -\frac{1}{b} \log\left(-\frac{b}{a} e^{ax} - bC\right).
\]
**Hint:** When solving for a variable, ensure you apply logarithmic properties correctly.
### Final Solution
The solution to the differential equation is:
\[
y = -\frac{1}{b} \log\left(-\frac{b}{a} e^{ax} - bC\right).
\]
