What is the solution of the differential equation \(\frac{{dx}}{{dy}} = \frac{{x\; + {\rm{\;}}y\; + {\rm{\;}}1}}{{x\; + {\rm{\;}}y\; - {\rm{\;}}1}}?\) Where c is the arbitrary constant
y – x + ln(x + y) = c
Solving Differential Equations: Using Substitution Method
Let's find the solution to the given differential equation \(\frac{{dx}}{{dy}} = \frac{{x\; + {\rm{\;}}y\; + {\rm{\;}}1}}{{x\; + {\rm{\;}}y\; - {\rm{\;}}1}}\).
The structure of this differential equation, where the numerator and denominator involve the same linear combination of \(x\) and \(y\) (\(x+y\)), suggests using a substitution method.
Applying Substitution for the Differential Equation
Let us introduce a new variable, say \(v\), such that \(v = x + y\). This substitution is chosen because the term \(x+y\) appears multiple times in the equation.
Now, we need to find \(\frac{dx}{dy}\) in terms of \(v\) and \(\frac{dv}{dy}\). Differentiating the substitution \(v = x + y\) with respect to \(y\), we get:
\(\frac{dv}{dy} = \frac{dx}{dy} + \frac{dy}{dy}\)
\(\frac{dv}{dy} = \frac{dx}{dy} + 1\)
From this, we can express \(\frac{dx}{dy}\) as:
\(\frac{dx}{dy} = \frac{dv}{dy} - 1\)
Substituting into the Differential Equation
Now, we replace \(\frac{dx}{dy}\) and the expressions involving \(x+y\) in the original differential equation with their equivalents in terms of \(v\):
Original equation: \(\frac{{dx}}{{dy}} = \frac{{x\; + {\rm{\;}}y\; + {\rm{\;}}1}}{{x\; + {\rm{\;}}y\; - {\rm{\;}}1}}\)
Substitute \(\frac{dx}{dy} = \frac{dv}{dy} - 1\) and \(x+y = v\):
\(\frac{dv}{dy} - 1 = \frac{v + 1}{v - 1}\)
Now, we isolate \(\frac{dv}{dy}\):
\(\frac{dv}{dy} = 1 + \frac{v + 1}{v - 1}\)
Combine the terms on the right-hand side by finding a common denominator:
\(\frac{dv}{dy} = \frac{v - 1}{v - 1} + \frac{v + 1}{v - 1}\)
\(\frac{dv}{dy} = \frac{(v - 1) + (v + 1)}{v - 1}\)
\(\frac{dv}{dy} = \frac{2v}{v - 1}\)
Separating Variables and Integrating
The equation \(\frac{dv}{dy} = \frac{2v}{v - 1}\) is now a variable separable differential equation. We can rearrange it to group terms involving \(v\) with \(dv\) and terms involving \(y\) with \(dy\):
\(\frac{v - 1}{2v} dv = dy\)
We can split the fraction on the left-hand side:
\(\left(\frac{v}{2v} - \frac{1}{2v}\right) dv = dy\)
\(\left(\frac{1}{2} - \frac{1}{2v}\right) dv = dy\)
Now, we integrate both sides of the equation:
\(\int \left(\frac{1}{2} - \frac{1}{2v}\right) dv = \int dy\)
Integrating the left side with respect to \(v\) and the right side with respect to \(y\):
\(\frac{1}{2}v - \frac{1}{2}\ln|v| = y + C'\) (where \(C'\) is the constant of integration)
Substituting Back to Original Variables
Now, we substitute back \(v = x + y\) into the integrated equation:
\(\frac{1}{2}(x + y) - \frac{1}{2}\ln|x + y| = y + C'\)
To simplify and rearrange, we can multiply the entire equation by 2:
\((x + y) - \ln|x + y| = 2y + 2C'\)
Now, move all terms to one side to match the form of the given options. Let's gather the \(y\) terms:
\(x + y - 2y - \ln|x + y| = 2C'\)
\(x - y - \ln|x + y| = 2C'\)
We can move the term with \(\ln\) to the other side or rearrange differently. Let's aim for the form \(y - x + \ln(x+y) = c\).
From \(x - y - \ln|x + y| = 2C'\), we can multiply by -1:
\(-(x - y - \ln|x + y|) = -(2C')\)
\(-x + y + \ln|x + y| = -2C'\)
Let the new constant \(-2C'\) be \(c\).
\(y - x + \ln|x + y| = c\)
Assuming \(x+y\) is such that \(\ln(x+y)\) is defined and the absolute value can be omitted, the solution is:
\(y - x + \ln(x + y) = c\)
Summary of Steps:
- Identify the structure of the differential equation.
- Apply the substitution \(v = x+y\).
- Differentiate the substitution with respect to \(y\) to find \(\frac{dx}{dy}\).
- Substitute into the original differential equation.
- Simplify the equation in terms of \(v\) and \(y\).
- Separate the variables \(v\) and \(y\).
- Integrate both sides.
- Substitute back \(v = x+y\) to get the solution in terms of \(x\) and \(y\).
- Rearrange the terms to match the options.
| Step | Description | Equation |
|---|---|---|
| 1 | Original Equation | \(\frac{{dx}}{{dy}} = \frac{{x\; + {\rm{\;}}y\; + {\rm{\;}}1}}{{x\; + {\rm{\;}}y\; - {\rm{\;}}1}}\) |
| 2 | Substitution | \(v = x + y\) |
| 3 | Derivative of Substitution | \(\frac{dx}{dy} = \frac{dv}{dy} - 1\) |
| 4 | Substitute & Simplify | \(\frac{dv}{dy} = \frac{2v}{v - 1}\) |
| 5 | Separate Variables | \(\frac{v - 1}{2v} dv = dy\) |
| 6 | Integrate | \(\frac{1}{2}v - \frac{1}{2}\ln|v| = y + C'\) |
| 7 | Substitute Back | \(\frac{1}{2}(x + y) - \frac{1}{2}\ln|x + y| = y + C'\) |
| 8 | Final Solution Form | \(y - x + \ln|x + y| = c\) |
Revision Table: Differential Equation Concepts
Understanding the different types of differential equations and methods to solve them is crucial.
- First-Order Differential Equation: An equation involving only the first derivative of the dependent variable.
- Variable Separable Method: A method where the equation can be rearranged so that all terms involving the dependent variable are on one side and all terms involving the independent variable are on the other side.
- Substitution Method: Used when a specific combination of variables appears repeatedly in the equation. Choosing the right substitution simplifies the equation.
- Constant of Integration: An arbitrary constant introduced when performing indefinite integration. It represents the family of solutions to the differential equation.
Additional Information: Homogeneous and Non-Homogeneous Equations
Differential equations can be classified based on their structure. The given differential equation, after the substitution \(v=x+y\), transforms into a variable separable form.
Some differential equations are classified as homogeneous or non-homogeneous. A first-order equation \(\frac{dy}{dx} = f(x, y)\) is homogeneous if \(f(tx, ty) = f(x, y)\) for some constant \(t\). The given equation is not directly homogeneous, but the form \(\frac{dx}{dy} = f(x+y)\) is a specific type solvable by the substitution \(v=x+y\).
The solution \(y - x + \ln(x + y) = c\) represents a family of curves. The specific value of the constant \(c\) depends on initial or boundary conditions, if provided.
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