What is the solution of (1 +2x) dy – (1 – 2y) dx = 0?
x – y – 2xy = c
Solving the Differential Equation by Separation of Variables
The given equation is a first-order differential equation: \( (1 + 2x) dy – (1 – 2y) dx = 0 \). We can solve this equation by separating the variables.
Separating Variables
First, rearrange the equation to group terms with \(dy\) and \(dx\):
\( (1 + 2x) dy = (1 – 2y) dx \)Now, move all terms involving \(y\) to the side with \(dy\) and all terms involving \(x\) to the side with \(dx\). Assuming \(1 - 2y \neq 0\) and \(1 + 2x \neq 0\), we can divide both sides:
\( \frac{dy}{1 – 2y} = \frac{dx}{1 + 2x} \)This form now has variables separated.
Integrating Both Sides
To find the general solution, integrate both sides of the separated equation:
\( \int \frac{dy}{1 – 2y} = \int \frac{dx}{1 + 2x} \)Let's evaluate each integral separately.
Left Side Integral
For the left side, \( \int \frac{dy}{1 – 2y} \), we can use a substitution. Let \(u = 1 – 2y\). Then, the differential \(du = -2 dy\), which implies \(dy = –\frac{1}{2} du\). Substituting this into the integral:
\( \int \frac{–\frac{1}{2} du}{u} = –\frac{1}{2} \int \frac{1}{u} du \)The integral of \(1/u\) is \( \ln|u| \). So, the left side integral becomes:
\( –\frac{1}{2} \ln|1 – 2y| + C_1 \)where \(C_1\) is the constant of integration.
Right Side Integral
For the right side, \( \int \frac{dx}{1 + 2x} \), we use a similar substitution. Let \(v = 1 + 2x\). Then, the differential \(dv = 2 dx\), which implies \(dx = \frac{1}{2} dv\). Substituting this into the integral:
\( \int \frac{\frac{1}{2} dv}{v} = \frac{1}{2} \int \frac{1}{v} dv \)The integral of \(1/v\) is \( \ln|v| \). So, the right side integral becomes:
\( \frac{1}{2} \ln|1 + 2x| + C_2 \)where \(C_2\) is the constant of integration.
Combining and Simplifying the Solution
Equating the results of the two integrals:
\( –\frac{1}{2} \ln|1 – 2y| + C_1 = \frac{1}{2} \ln|1 + 2x| + C_2 \)Combine the constants into a single constant \(C' = C_2 – C_1\):
\( –\frac{1}{2} \ln|1 – 2y| = \frac{1}{2} \ln|1 + 2x| + C' \)Multiply by 2 to remove the fractions:
\( –\ln|1 – 2y| = \ln|1 + 2x| + 2C' \)Move the log terms to one side:
\( 0 = \ln|1 + 2x| + \ln|1 – 2y| + 2C' \)Using the logarithm property \( \ln A + \ln B = \ln(AB) \):
\( \ln| (1 + 2x)(1 – 2y) | = –2C' \)Exponentiate both sides with base \(e\):
\( | (1 + 2x)(1 – 2y) | = e^{–2C'} \)Let \(C = \pm e^{–2C'}\). \(C\) is an arbitrary non-zero constant. If we also consider the cases where \(1-2y=0\) or \(1+2x=0\), which are constant solutions \(y=1/2\) and \(x=-1/2\), these can sometimes be included in the general solution form with \(C=0\). So, \(C\) can be any arbitrary constant.
\( (1 + 2x)(1 – 2y) = C \)Now, expand the left side of the equation:
\( 1(1 – 2y) + 2x(1 – 2y) = C \) \( 1 – 2y + 2x – 4xy = C \)Rearrange the terms to match the form of the options:
\( 2x – 2y – 4xy = C – 1 \)Divide the entire equation by 2 (note that \( (C – 1)/2 \) is just another arbitrary constant):
\( x – y – 2xy = \frac{C – 1}{2} \)Let the new arbitrary constant be denoted by \(c\), where \( c = \frac{C – 1}{2} \).
The general solution is:
\( x – y – 2xy = c \)This matches one of the provided options.
Summary of Solution Steps
| Step | Action | Result |
|---|---|---|
| 1 | Separate variables | \( \frac{dy}{1 – 2y} = \frac{dx}{1 + 2x} \) |
| 2 | Integrate both sides | \( \int \frac{dy}{1 – 2y} = \int \frac{dx}{1 + 2x} \) |
| 3 | Evaluate integrals | \( –\frac{1}{2} \ln|1 – 2y| = \frac{1}{2} \ln|1 + 2x| + C' \) |
| 4 | Simplify and exponentiate | \( (1 + 2x)(1 – 2y) = C \) |
| 5 | Expand and rearrange | \( x – y – 2xy = c \) |
Revision Table: Key Differential Equation Concepts
| Concept | Description | Application Here |
|---|---|---|
| Differential Equation | An equation involving a function and its derivatives. | \( (1 + 2x) dy – (1 – 2y) dx = 0 \) is a first-order DE. |
| First-Order DE | Involves only the first derivative of the function. | Our DE has only \(dy/dx\). |
| Separable DE | A first-order DE that can be written in the form \( N(y) dy = M(x) dx \). | We successfully transformed the given DE into this form. |
| General Solution | A solution that contains arbitrary constants, covering all possible specific solutions. | \( x – y – 2xy = c \) is the general solution. |
| Integration | The process of finding the antiderivative, used here to reverse differentiation and solve the DE. | Used to integrate \( \frac{dy}{1 – 2y} \) and \( \frac{dx}{1 + 2x} \). |
Additional Information: Separation of Variables Method
The method of separation of variables is a powerful technique for solving first-order differential equations that can be written in the form \( \frac{dy}{dx} = f(x) g(y) \). The steps are:
- Rewrite the equation as \( \frac{dy}{g(y)} = f(x) dx \). Note that this step requires \( g(y) \neq 0 \).
- Integrate both sides of the separated equation: \( \int \frac{dy}{g(y)} = \int f(x) dx \).
- Solve the integrals to obtain an equation involving \(x\), \(y\), and an arbitrary constant of integration.
- Simplify the resulting equation to express the general solution, often in an implicit form \( H(x, y) = C \) or sometimes explicitly as \( y = \phi(x) \).
It's important to check for any constant solutions that might have been excluded when dividing by \(g(y)\).
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