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(x-2y+5)dx+(2x-y+4)dy=0 Solution

4 min read Jun 17, 2024
(x-2y+5)dx+(2x-y+4)dy=0 Solution

Solving the Differential Equation (x-2y+5)dx + (2x-y+4)dy = 0

This article will guide you through the process of finding the general solution to the given differential equation:

(x - 2y + 5)dx + (2x - y + 4)dy = 0

Understanding the Equation

This is a first-order non-exact differential equation. To solve it, we will follow these steps:

  1. Check for exactness:

    • We need to see if the equation is in the form M(x,y)dx + N(x,y)dy = 0, where ∂M/∂y = ∂N/∂x.
    • In our case, M(x,y) = x - 2y + 5 and N(x,y) = 2x - y + 4.
    • ∂M/∂y = -2 and ∂N/∂x = 2. Since these are not equal, the equation is not exact.

  2. Find an integrating factor:

    • We need to find a function μ(x,y) such that multiplying both sides of the equation by μ(x,y) makes it exact.
    • In this case, we can use the following formula for μ:
    μ(x,y) = exp(∫(∂N/∂x - ∂M/∂y) / (M) dx)
    • Substituting the values:
      • μ(x,y) = exp(∫(2 - (-2)) / (x - 2y + 5) dx)
      • μ(x,y) = exp(∫4 / (x - 2y + 5) dx)
      • μ(x,y) = exp(4ln|x - 2y + 5|)
      • μ(x,y) = (x - 2y + 5)^4

  3. Multiply the equation by the integrating factor:

    • Multiplying the original equation by μ(x,y) = (x - 2y + 5)^4 gives us:
    • (x - 2y + 5)^5 dx + (2x - y + 4)(x - 2y + 5)^4 dy = 0

  4. Check for exactness again:

    • Now, ∂M/∂y = -10(x - 2y + 5)^4 = ∂N/∂x. The equation is now exact.

  5. Solve the exact equation:

    • We know that there exists a function φ(x,y) such that ∂φ/∂x = M and ∂φ/∂y = N.
    • Integrating ∂φ/∂x = M with respect to x, we get:
    φ(x,y) = (1/6)(x - 2y + 5)^6 + g(y)
    • where g(y) is an arbitrary function of y.
    • Differentiating φ(x,y) with respect to y and equating it to N, we get:
      • ∂φ/∂y = -2(x - 2y + 5)^5 + g'(y) = (2x - y + 4)(x - 2y + 5)^4
      • g'(y) = (2x - y + 4)(x - 2y + 5)^4 + 2(x - 2y + 5)^5
    • Solving for g(y), we get:
      • g(y) = (1/6)(x - 2y + 5)^6 + (1/3)(x - 2y + 5)^5 + C
      • Where C is an arbitrary constant.

  6. General Solution:

    • Substituting the value of g(y) in φ(x,y), we get the general solution:
    φ(x,y) = (1/3)(x - 2y + 5)^6 + C = 0

Conclusion

The general solution to the differential equation (x - 2y + 5)dx + (2x - y + 4)dy = 0 is:

(1/3)(x - 2y + 5)^6 + C = 0

Where C is an arbitrary constant. This solution represents a family of curves, each defined by a specific value of C.

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