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

4 min read Jun 17, 2024

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

This article will delve into the solution of the given differential equation:

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

We will use the method of exact differential equations to find the general solution.

Before we can proceed, we need to verify if the given equation is indeed an exact differential equation.

Definition: A differential equation of the form M(x, y)dx + N(x, y)dy = 0 is called exact if and only if

∂M/∂y = ∂N/∂x.

In our case, M(x, y) = x^2y - 2xy^2 and N(x, y) = -x^3 + 3x^2y. Let's calculate the partial derivatives:

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

To solve this, we can try to find an integrating factor μ(x, y) such that when we multiply the equation by μ, it becomes exact.

Let's consider the integrating factor μ to be a function of x only, i.e., μ = μ(x). Multiplying the equation by μ, we get:

(μx^2y - 2μxy^2)dx - (μx^3 - 3μx^2y)dy = 0

Now, we need to ensure that this new equation is exact. This means:

∂(μx^2y - 2μxy^2)/∂y = ∂(-μx^3 + 3μx^2y)/∂x

This simplifies to:

μx^2 - 4μxy = -3μx^2 + 6μxy

Solving for μ, we get:

μ = 1/x^2

Therefore, the integrating factor is 1/x^2.

Now, multiply the original equation by the integrating factor 1/x^2:

(y - 2y^2/x)dx - (x - 3y)dy = 0

This equation is now exact. Let's find the solution.

Step 1: Find the potential function F(x, y) such that:

∂F/∂x = y - 2y^2/x ∂F/∂y = -x + 3y

Step 2: Integrate the first equation with respect to x:

F(x, y) = xy - 2y^2ln|x| + g(y)

Step 3: Differentiate the result with respect to y and equate it to the second equation:

∂F/∂y = x - 4yln|x| + g'(y) = -x + 3y

This implies g'(y) = 7y, and hence g(y) = (7/2)y^2 + C.

Step 4: Substitute the value of g(y) back into F(x, y):

F(x, y) = xy - 2y^2ln|x| + (7/2)y^2 + C

Therefore, the general solution of the given differential equation is:

xy - 2y^2ln|x| + (7/2)y^2 = C

where C is an arbitrary constant.