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

Jasonbradley

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Jun 17, 2024

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Solving the Differential Equation: (x^3 + 3xy^2)dx + (y^3 + 3x^2y)dy = 0

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

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

This equation is a first-order homogeneous differential equation, meaning it can be expressed in the form:

M(x, y) dx + N(x, y) dy = 0

where M and N are homogeneous functions of the same degree.

1. Identifying Homogeneity

Let's verify if our equation is homogeneous.

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

We can see that both M and N are homogeneous functions of degree 3. This is because replacing x and y with kx and ky, respectively, will result in k^3 multiplied by the original function:

• M(kx, ky) = (kx)^3 + 3(kx)(ky)^2 = k^3(x^3 + 3xy^2) = k^3M(x, y)
• N(kx, ky) = (ky)^3 + 3(kx)^2(ky) = k^3(y^3 + 3x^2y) = k^3N(x, y)

Therefore, the given differential equation is indeed homogeneous.

2. Solving the Homogeneous Equation

To solve this type of equation, we introduce a new variable:

v = y/x

This leads to:

• y = vx
• dy = vdx + xdv

Substituting these values into the original equation:

(x^3 + 3x(vx)^2)dx + ((vx)^3 + 3x^2(vx))(vdx + xdv) = 0

Simplifying the equation:

(x^3 + 3x^3v^2)dx + (v^3x^3 + 3x^3v)(vdx + xdv) = 0

x^3(1 + 3v^2)dx + x^3(v^3 + 3v)(vdx + xdv) = 0

Dividing both sides by x^3 and rearranging terms:

(1 + 3v^2 + v^4 + 3v^2)dx + (v^4 + 3v^2)xdv = 0

(1 + 6v^2 + v^4)dx + (v^4 + 3v^2)xdv = 0

Now, we can separate the variables:

(1 + 6v^2 + v^4) / (v^4 + 3v^2) dv = -dx/x

Integrating both sides:

∫ (1 + 6v^2 + v^4) / (v^4 + 3v^2) dv = -∫ dx/x

This integral can be solved using partial fractions, leading to:

ln|v| - 2ln|v^2 + 3| = -ln|x| + C

Where C is the constant of integration.

3. Expressing the Solution in terms of x and y

Recall that v = y/x. Substitute this back into the equation and simplify:

ln|y/x| - 2ln|(y/x)^2 + 3| = -ln|x| + C

ln|y| - ln|x| - 2ln|(y^2 + 3x^2)/x^2| = -ln|x| + C

ln|y| - 2ln|y^2 + 3x^2| + 2ln|x^2| = C

ln|y| - 2ln|y^2 + 3x^2| + 4ln|x| = C

This is the implicit general solution to the given differential equation. The solution can be expressed in various forms depending on the context and desired level of simplification.

Note: This solution represents a family of curves. A specific solution can be found by applying an initial condition (a point that lies on the curve).

This analysis demonstrates the process of solving a homogeneous first-order differential equation using the substitution method.