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(y^4+2y)dx+(xy^3+2y^4-4x)dy=0

4 min read Jun 17, 2024
(y^4+2y)dx+(xy^3+2y^4-4x)dy=0

Solving the Differential Equation (y⁴+2y)dx + (xy³+2y⁴-4x)dy = 0

This article will guide you through solving the given differential equation:

(y⁴+2y)dx + (xy³+2y⁴-4x)dy = 0

We will use the following steps:

  1. Identifying the type of differential equation: The given equation is a first-order, non-linear, exact differential equation.

  2. Checking for exactness: 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) = y⁴+2y
    • N(x,y) = xy³+2y⁴-4x

    Calculating the partial derivatives:

    • ∂M/∂y = 4y³+2
    • ∂N/∂x = y³-4

    Since ∂M/∂y ≠ ∂N/∂x, the given equation is not exact.

  3. Finding an integrating factor: To make the equation exact, we need to find an integrating factor μ(x,y) such that:

    • μ(x,y)M(x,y)dx + μ(x,y)N(x,y)dy = 0 is exact.
    • ∂(μM)/∂y = ∂(μN)/∂x

    In this case, we can find an integrating factor that depends only on y, μ(y). Let's find this integrating factor:

    • ∂(μM)/∂y = μ(4y³+2) + μ'(y)(y⁴+2y)
    • ∂(μN)/∂x = μ(y³-4)

    Equating the two, we get:

    • μ(4y³+2) + μ'(y)(y⁴+2y) = μ(y³-4)

    Simplifying the equation:

    • μ'(y)(y⁴+2y) = -6μy³ - 6μ

    Dividing both sides by μ(y⁴+2y):

    • μ'(y)/μ = -6y³/(y⁴+2y)

    Integrating both sides:

    • ln|μ| = -3ln|y² + 2| + C
    • μ = e^(-3ln|y² + 2|) = 1/(y² + 2)³

  4. Multiplying by the integrating factor: We multiply the original equation by the integrating factor 1/(y² + 2)³:

    • (y⁴+2y)/(y² + 2)³ dx + (xy³+2y⁴-4x)/(y² + 2)³ dy = 0

  5. Verifying exactness: Now, we need to check if the equation is exact:

    • ∂M/∂y = ∂[(y⁴+2y)/(y² + 2)³]/∂y = 0
    • ∂N/∂x = ∂[(xy³+2y⁴-4x)/(y² + 2)³]/∂x = 0

    Since ∂M/∂y = ∂N/∂x, the equation is now exact.

  6. Solving the exact equation: To solve the exact equation, we find a potential function Φ(x,y) such that:

    • ∂Φ/∂x = M(x,y) and ∂Φ/∂y = N(x,y)

    Integrating ∂Φ/∂x = M(x,y) with respect to x:

    • Φ(x,y) = ∫(y⁴+2y)/(y² + 2)³ dx + g(y)
    • Φ(x,y) = -x/(y² + 2)² + g(y)

    Differentiating Φ(x,y) with respect to y and equating it to N(x,y):

    • ∂Φ/∂y = 4xy/(y² + 2)³ + g'(y) = (xy³+2y⁴-4x)/(y² + 2)³

    Solving for g'(y):

    • g'(y) = -2y⁴/(y² + 2)³

    Integrating g'(y) with respect to y:

    • g(y) = y/(y² + 2)² + C

    Therefore, the general solution to the differential equation is:

    • Φ(x,y) = -x/(y² + 2)² + y/(y² + 2)² + C = 0
    • -x + y + C(y² + 2)² = 0

This represents the general solution to the given differential equation.

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