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

Jasonbradley

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Jun 16, 2024

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Solving the Differential Equation (3x²y + 2xy + y³)dx + (x² + y²)dy = 0

This article explores the solution to the given differential equation:

(3x²y + 2xy + y³)dx + (x² + y²)dy = 0

This equation falls under the category of exact differential equations. To determine if an equation is exact, we need to check if it satisfies the following condition:

∂M/∂y = ∂N/∂x

Where:

• M(x, y) = 3x²y + 2xy + y³ (the coefficient of dx)
• N(x, y) = x² + y² (the coefficient of dy)

Let's calculate the partial derivatives:

• ∂M/∂y = 3x² + 2x + 3y²
• ∂N/∂x = 2x

Since ∂M/∂y ≠ ∂N/∂x, the equation is not exact in its current form. To solve this, we need to find an integrating factor that will make the equation exact.

Finding the Integrating Factor

An integrating factor is a function µ(x, y) that when multiplied by the original equation, will transform it into an exact equation.

We can use the following formula to find the integrating factor:

µ(x, y) = exp(∫(∂N/∂x - ∂M/∂y)/M dx)

Let's calculate:

• (∂N/∂x - ∂M/∂y)/M = (2x - (3x² + 2x + 3y²))/(3x²y + 2xy + y³) = (-3x² - 3y²)/(3x²y + 2xy + y³)

To simplify, we can factor out -3:

• (-3x² - 3y²)/(3x²y + 2xy + y³) = -3(x² + y²)/(y(3x² + 2x + y²))

Now, integrating this expression with respect to x, we get:

• ∫(-3(x² + y²)/(y(3x² + 2x + y²))) dx = -ln(y) + C(y)

where C(y) is an arbitrary function of y.

Since we are looking for a function that depends only on y, we can discard the constant of integration C(y). Therefore, our integrating factor is:

µ(y) = exp(-ln(y)) = 1/y

Solving the Exact Equation

Now, we multiply the original equation by the integrating factor:

((3x²y + 2xy + y³)/y)dx + ((x² + y²)/y)dy = 0

Simplifying:

(3x² + 2x + y²)dx + (x²/y + y)dy = 0

This equation is now exact because:

• ∂(3x² + 2x + y²)/∂y = 2y
• ∂(x²/y + y)/∂x = 2x/y

To solve this exact equation, we follow these steps:

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

   • ∂F/∂x = M(x, y) = 3x² + 2x + y²
   • ∂F/∂y = N(x, y) = x²/y + y

2. Integrate ∂F/∂x with respect to x:

   • F(x, y) = ∫(3x² + 2x + y²)dx = x³ + x² + xy² + g(y)

   where g(y) is an arbitrary function of y.

3. Differentiate F(x, y) with respect to y and equate it to N(x, y):

   • ∂F/∂y = 2xy + g'(y) = x²/y + y
   • g'(y) = y
   • g(y) = y²/2 + C

4. Substitute g(y) back into F(x, y):

   • F(x, y) = x³ + x² + xy² + y²/2 + C

Therefore, the general solution to the original differential equation is:

x³ + x² + xy² + y²/2 = C

This solution represents a family of curves. Each curve corresponds to a specific value of the constant C.