What is the solution of the differential equation \(\ln \left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right) = {\rm{ax}} + {\rm{by}}?\)
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
| Step | Action | Purpose |
|---|---|---|
| 1 | Identify the type of differential equation. | Determine the appropriate solution method. |
| 2 | Separate variables (if possible). | Rearrange the equation so each side contains only one variable and its differential. |
| 3 | Integrate both sides. | Apply the integral operator to both sides of the separated equation. |
| 4 | Evaluate integrals. | Find the antiderivatives of the expressions on both sides. |
| 5 | Solve for the dependent variable (if required) and include the constant of integration. | Express the solution clearly and account for the arbitrary constant introduced during integration. |
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.
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