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Solving the Differential Equation (2x-4y+5)dy+(x-2y+3)dx=0
This article explores the solution of the given differential equation:
(2x-4y+5)dy+(x-2y+3)dx=0
This equation is a first-order, non-linear differential equation. To solve it, we will use the following steps:
1. Check for Exactness
The first step is to check if the equation is exact. A differential equation of the form M(x, y)dx + N(x, y)dy = 0 is exact if:
∂M/∂y = ∂N/∂x
In our case:
- M(x, y) = x - 2y + 3
- N(x, y) = 2x - 4y + 5
Calculating the partial derivatives:
- ∂M/∂y = -2
- ∂N/∂x = 2
Since ∂M/∂y ≠ ∂N/∂x, the equation is not exact.
2. Finding an Integrating Factor
To make the equation exact, we need to find an integrating factor, which is a function that, when multiplied by the original equation, makes it exact.
We can find the integrating factor by using the following formulas:
- If (∂M/∂y - ∂N/∂x)/N is a function of x only, then the integrating factor is e^(∫(∂M/∂y - ∂N/∂x)/N dx)
- If (∂N/∂x - ∂M/∂y)/M is a function of y only, then the integrating factor is e^(∫(∂N/∂x - ∂M/∂y)/M dy)
In our case:
- (∂M/∂y - ∂N/∂x)/N = (-2 - 2)/(2x - 4y + 5) = -4/(2x - 4y + 5), which is not a function of x only.
- (∂N/∂x - ∂M/∂y)/M = (2 + 2)/(x - 2y + 3) = 4/(x - 2y + 3), which is not a function of y only.
Therefore, we need to find a different approach to find an integrating factor. One common approach is to try an integrating factor of the form:
μ(x, y) = e^(∫(P(x)dx + Q(y)dy))
Where P(x) and Q(y) are functions of x and y respectively.
By substituting the values of M and N in the equation (2x-4y+5)dy+(x-2y+3)dx=0 and comparing with the standard form M(x, y)dx + N(x, y)dy = 0, we can find that:
- P(x) = 1/(x - 2y + 3)
- Q(y) = -1/(2x - 4y + 5)
Then, the integrating factor will be:
μ(x, y) = e^(∫(1/(x - 2y + 3)dx - 1/(2x - 4y + 5)dy))
It is difficult to directly integrate this expression. Therefore, we can try a simpler integrating factor of the form:
μ(x, y) = x^m * y^n
Where m and n are constants. We can substitute this into the equation and solve for m and n to make the equation exact.
3. Solving the Exact Equation
After finding the integrating factor, we multiply the original equation by it. This will result in an exact differential equation. We can then solve this equation by following these steps:
- Find the potential function F(x, y) such that ∂F/∂x = M and ∂F/∂y = N.
- The general solution to the differential equation is given by F(x, y) = C, where C is an arbitrary constant.
4. General Solution
By following these steps, we can find the general solution to the differential equation. This solution will be in the form of an implicit equation.
Conclusion
The solution process for this differential equation is complex and requires multiple steps. Finding an appropriate integrating factor is crucial, and different approaches can be explored. Once the equation is made exact, the solution can be obtained through finding a potential function. Remember that the final solution will be an implicit equation.