dx%2B(2y-x-1)dy%3D0.jpeg "(2x-y+1)dx+(2y-x-1)dy=0")
Solving the Differential Equation (2x-y+1)dx + (2y-x-1)dy = 0
This article will guide you through the process of solving the given differential equation:
(2x-y+1)dx + (2y-x-1)dy = 0
This is a first-order differential equation that is not exact. To solve it, we will use an integrating factor.
1. Checking for Exactness
A differential equation in the form M(x, y)dx + N(x, y)dy = 0 is exact if:
∂M/∂y = ∂N/∂x
In our case, we have:
- M(x, y) = 2x - y + 1
- N(x, y) = 2y - x - 1
Calculating the partial derivatives:
- ∂M/∂y = -1
- ∂N/∂x = -1
Since ∂M/∂y = ∂N/∂x, the equation is exact.
2. Finding the Integrating Factor
Since the equation is not exact, we need to find an integrating factor, denoted by μ(x, y). To find μ(x, y), we can use the following steps:
- Calculate (∂N/∂x - ∂M/∂y)/M: In our case, this equals 0/M = 0.
- Since the result is a function of y only, the integrating factor is μ(y) = exp(∫(∂N/∂x - ∂M/∂y)/M dy).
- Therefore, μ(y) = exp(∫0 dy) = exp(C) = C, where C is a constant.
For simplicity, we can choose C = 1.
3. Multiplying the Equation by the Integrating Factor
We multiply the original equation by the integrating factor:
(2x-y+1)dx + (2y-x-1)dy = 0 ⇒ (2x-y+1)dx + (2y-x-1)dy = 0
4. Solving the Exact Equation
Since the equation is now exact, we can find a solution by integrating M(x, y) with respect to x and then differentiating the result with respect to y to find a term that can be added to make it equal to N(x, y).
- ∫M(x, y) dx = ∫(2x-y+1) dx = x² - xy + x + C(y)
- ∂/∂y (x² - xy + x + C(y)) = -x + C'(y)
- To make this equal to N(x, y), we need C'(y) = 2y - 1.
- Integrating this, we get C(y) = y² - y + K, where K is a constant.
Therefore, the general solution to the differential equation is:
x² - xy + x + y² - y + K = 0
Conclusion
We have successfully solved the differential equation (2x-y+1)dx + (2y-x-1)dy = 0 by using an integrating factor. The general solution is x² - xy + x + y² - y + K = 0, where K is a constant.