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 \(g(y)\).
Important Questions from Differential Equations
What is the differential equation of all parabolas of the type y2 = 4a (x - b)?
What is the order of the differential equation ?
What is the degree of the differential equation ?
A solution of the differential equation
\(\left(\frac{d y}{d x}\right)^2-x \frac{d y}{d x}=0 \) is
If x dy = y dx + y 2dy, y > 0 and y (1) = 1, then what is y (-3) equal to?
