Solving the Differential Equation \(\ln \left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right) = {\rm{ax}} + {\rm{by}}\)

We are asked to find the solution to the given first-order differential equation:

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

This equation involves a logarithm. To solve it, we need to isolate the \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}}\) term. We can do this by exponentiating both sides of the equation with base \(e\).

Step 1: Exponentiate both sides

Taking the exponential of both sides of the equation \( \ln \left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right) = {\rm{ax}} + {\rm{by}} \), we get:

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

Using the property \(e^{\ln(u)} = u\), the left side simplifies:

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

Now, we can use the property of exponents \(e^{m+n} = e^m \cdot e^n\) to separate the terms on the right side:

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

Step 2: Separate the Variables

The equation is now in a form where we can separate the variables \(x\) and \(y\). We want all terms involving \(y\) (including \({\rm{dy}}\)) on one side and all terms involving \(x\) (including \({\rm{dx}}\)) on the other side.

Divide both sides by \(e^{{\rm{by}}}\) and multiply both sides by \({\rm{dx}}\):

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

We can rewrite the term on the left side using the property \(\frac{1}{e^u} = e^{-u}\):

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

The variables are now separated, with \(y\) terms on the left and \(x\) terms on the right.

Step 3: Integrate Both Sides

To find the solution to the differential equation, we integrate both sides of the separated equation:

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

Step 4: Perform the Integration

We need to evaluate the integrals. The general integral of \(e^{ku}\) with respect to \(u\) is \(\frac{1}{k}e^{ku} + C\), where \(C\) is the constant of integration.

For the left side integral, \(\int e^{-{\rm{by}}} \, {\rm{dy}}\), \(u=y\) and \(k=-b\). The integral is:

\[ \int e^{-{\rm{by}}} \, {\rm{dy}} = \frac{1}{-b} e^{-{\rm{by}}} + C_1 = -\frac{1}{b} e^{-{\rm{by}}} + C_1 \]

For the right side integral, \(\int e^{{\rm{ax}}} \, {\rm{dx}}\), \(u=x\) and \(k=a\). The integral is:

\[ \int e^{{\rm{ax}}} \, {\rm{dx}} = \frac{1}{a} e^{{\rm{ax}}} + C_2 \]

Equating the results of the integration:

\[ -\frac{1}{b} e^{-{\rm{by}}} + C_1 = \frac{1}{a} e^{{\rm{ax}}} + C_2 \]

Step 5: Rearrange and Simplify

We can rearrange the equation to group the exponential terms and combine the constants of integration \(C_1\) and \(C_2\) into a single constant \(C\).

Move the term \(-\frac{1}{b} e^{-{\rm{by}}}\) to the right side:

\[ C_1 = \frac{1}{a} e^{{\rm{ax}}} + \frac{1}{b} e^{-{\rm{by}}} + C_2 \]

Move \(C_2\) to the left side and define \(C = C_1 - C_2\):

\[ C_1 - C_2 = \frac{1}{a} e^{{\rm{ax}}} + \frac{1}{b} e^{-{\rm{by}}} \]

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

Or, written in a standard form:

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

This is the general solution to the differential equation.

Revision Table: Steps to Solve Differential Equations

Additional Information on Variable Separable Differential Equations

A first-order differential equation of the form \(\frac{{\rm{dy}}}{{\rm{dx}}} = f(x, y)\) is called variable separable if the function \(f(x, y)\) can be written as a product of a function of \(x\) only and a function of \(y\) only. That is, \(f(x, y) = g(x)h(y)\).

The equation can then be written as:

\[ \frac{{\rm{dy}}}{{\rm{dx}}} = g(x)h(y) \]

If \(h(y) \neq 0\), we can separate the variables by rearranging the equation:

\[ \frac{1}{h(y)} \, {\rm{dy}} = g(x) \, {\rm{dx}} \]

This form allows us to integrate each side with respect to its respective variable:

\[ \int \frac{1}{h(y)} \, {\rm{dy}} = \int g(x) \, {\rm{dx}} \]

Evaluating these integrals gives the general solution in the form \(H(y) = G(x) + C\), where \(H(y)\) is the antiderivative of \(1/h(y)\), \(G(x)\) is the antiderivative of \(g(x)\), and \(C\) is the constant of integration.

In our problem, \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = e^{{\rm{ax}}} \cdot e^{{\rm{by}}}\), where \(g(x) = e^{{\rm{ax}}}\) and \(h(y) = e^{{\rm{by}}}\). This confirms it is a variable separable differential equation.