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(x^(3)y^(3)+x^(2)y^(2)+xy+1)ydx+(x^(3)y^(3)-x^(2)y^(2)-xy+1)xdy=0

6 min read Jun 17, 2024
(x^(3)y^(3)+x^(2)y^(2)+xy+1)ydx+(x^(3)y^(3)-x^(2)y^(2)-xy+1)xdy=0

Solving the Differential Equation: (x^(3)y^(3)+x^(2)y^(2)+xy+1)ydx+(x^(3)y^(3)-x^(2)y^(2)-xy+1)xdy=0

This article will explore the solution of the given differential equation:

(x^(3)y^(3)+x^(2)y^(2)+xy+1)ydx+(x^(3)y^(3)-x^(2)y^(2)-xy+1)xdy=0

This differential equation is classified as exact because it satisfies the following condition:

∂M/∂y = ∂N/∂x

Where:

  • M = (x^(3)y^(3)+x^(2)y^(2)+xy+1)y
  • N = (x^(3)y^(3)-x^(2)y^(2)-xy+1)x

Let's calculate the partial derivatives:

  • ∂M/∂y = 3x^(3)y^(2) + 2x^(2)y + x + 1
  • ∂N/∂x = 3x^(2)y^(3) - 2xy^(2) - y + 1

We can see that ∂M/∂y = ∂N/∂x, confirming the equation is exact.

Solving the Exact Differential Equation

To solve an exact differential equation, we follow these steps:

  1. Find a function Ψ(x, y) such that:

    • ∂Ψ/∂x = M
    • ∂Ψ/∂y = N

  2. Integrate ∂Ψ/∂x = M with respect to x, treating y as a constant:

    • Ψ(x, y) = ∫M dx + h(y)
    • where h(y) is an arbitrary function of y.

  3. Differentiate Ψ(x, y) with respect to y:

    • ∂Ψ/∂y = ∂/∂y [∫M dx + h(y)]

  4. Equate the result to N and solve for h'(y):

    • ∂Ψ/∂y = N
    • Solve for h'(y)

  5. Integrate h'(y) to find h(y):

    • h(y) = ∫h'(y) dy

  6. Substitute h(y) back into Ψ(x, y):

    • Ψ(x, y) = ∫M dx + h(y)

  7. The general solution is given by:

    • Ψ(x, y) = C
    • where C is an arbitrary constant.

Applying the Steps to our Equation

  1. Find Ψ(x, y):

    • ∂Ψ/∂x = (x^(3)y^(3)+x^(2)y^(2)+xy+1)y
    • ∂Ψ/∂y = (x^(3)y^(3)-x^(2)y^(2)-xy+1)x

  2. Integrate ∂Ψ/∂x with respect to x:

    • Ψ(x, y) = ∫(x^(3)y^(3)+x^(2)y^(2)+xy+1)y dx + h(y)
    • Ψ(x, y) = (1/4)x^(4)y^(3) + (1/3)x^(3)y^(2) + (1/2)x^(2)y + xy + h(y)

  3. Differentiate Ψ(x, y) with respect to y:

    • ∂Ψ/∂y = (3/4)x^(4)y^(2) + (2/3)x^(3)y + (1/2)x^(2) + x + h'(y)

  4. Equate ∂Ψ/∂y to N and solve for h'(y):

    • (3/4)x^(4)y^(2) + (2/3)x^(3)y + (1/2)x^(2) + x + h'(y) = (x^(3)y^(3)-x^(2)y^(2)-xy+1)x
    • h'(y) = 0

  5. Integrate h'(y):

    • h(y) = C_1
    • where C_1 is an arbitrary constant.

  6. Substitute h(y) back into Ψ(x, y):

    • Ψ(x, y) = (1/4)x^(4)y^(3) + (1/3)x^(3)y^(2) + (1/2)x^(2)y + xy + C_1

  7. The general solution is:

    • (1/4)x^(4)y^(3) + (1/3)x^(3)y^(2) + (1/2)x^(2)y + xy + C_1 = C
    • (1/4)x^(4)y^(3) + (1/3)x^(3)y^(2) + (1/2)x^(2)y + xy = C_2
    • where C_2 = C - C_1 is another arbitrary constant.

Therefore, the general solution to the differential equation (x^(3)y^(3)+x^(2)y^(2)+xy+1)ydx+(x^(3)y^(3)-x^(2)y^(2)-xy+1)xdy=0 is:

(1/4)x^(4)y^(3) + (1/3)x^(3)y^(2) + (1/2)x^(2)y + xy = C_2

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