(x^2+y^2+1)dx+x(x-2y)dy=0

6 min read Jun 17, 2024

Solving the Differential Equation: (x^2+y^2+1)dx + x(x-2y)dy = 0

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

(x^2+y^2+1)dx + x(x-2y)dy = 0

This equation is a non-exact first-order differential equation, meaning it cannot be directly solved by integrating both sides. However, we can use an integrating factor to make it exact.

1. Identifying the Integrating Factor

To find the integrating factor, we need to check if the equation satisfies the following condition:

(∂M/∂y) = (∂N/∂x)

Where:

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

Let's calculate the partial derivatives:

∂M/∂y = 2y
∂N/∂x = 2x - 2y

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

Now, let's find the integrating factor μ(x,y):

μ(x,y) = e^(∫[(∂N/∂x) - (∂M/∂y)]/M dx)

Plugging in the values:

μ(x,y) = e^(∫[(2x - 2y) - (2y)]/(x^2+y^2+1) dx)
μ(x,y) = e^(∫(2x - 4y)/(x^2+y^2+1) dx)

We can simplify this integral by splitting it into two parts:

μ(x,y) = e^(∫(2x)/(x^2+y^2+1) dx - ∫(4y)/(x^2+y^2+1) dx)

Solving the first integral, we get:

∫(2x)/(x^2+y^2+1) dx = ln(x^2+y^2+1)

The second integral cannot be solved directly. However, since it only involves 'y', we can treat it as a constant for this integration.

Therefore, the integrating factor is:

μ(x,y) = e^(ln(x^2+y^2+1) - (4y/x^2+y^2+1))
μ(x,y) = (x^2+y^2+1)e^(-4y/(x^2+y^2+1))

2. Making the Equation Exact

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

(x^2+y^2+1)e^(-4y/(x^2+y^2+1)) [(x^2+y^2+1)dx + x(x-2y)dy] = 0

Simplifying, we get:

(x^2+y^2+1)^2 e^(-4y/(x^2+y^2+1)) dx + x(x-2y)(x^2+y^2+1)e^(-4y/(x^2+y^2+1)) dy = 0

Now, the equation is exact because:

∂/∂y [(x^2+y^2+1)^2 e^(-4y/(x^2+y^2+1))] = ∂/∂x [x(x-2y)(x^2+y^2+1)e^(-4y/(x^2+y^2+1))]

3. Solving the Exact Equation

Since the equation is exact, it can be written as the total differential of a function u(x,y):

du = (∂u/∂x)dx + (∂u/∂y)dy

Comparing this to our exact equation, we get:

(∂u/∂x) = (x^2+y^2+1)^2 e^(-4y/(x^2+y^2+1))
(∂u/∂y) = x(x-2y)(x^2+y^2+1)e^(-4y/(x^2+y^2+1))

Integrating the first equation with respect to 'x', we get:

u(x,y) = ∫(x^2+y^2+1)^2 e^(-4y/(x^2+y^2+1)) dx + h(y)

Where h(y) is an arbitrary function of 'y'.

To find h(y), we differentiate this expression with respect to 'y' and compare it to the second equation:

∂u/∂y = ∂/∂y [∫(x^2+y^2+1)^2 e^(-4y/(x^2+y^2+1)) dx + h(y)]

After performing the differentiation and comparing with the second equation, we find:

h'(y) = 0

Therefore, h(y) is a constant.

4. The General Solution

Finally, the general solution to the given differential equation is:

u(x,y) = ∫(x^2+y^2+1)^2 e^(-4y/(x^2+y^2+1)) dx + C

Where C is an arbitrary constant.

This integral is complex and cannot be solved analytically. However, we have successfully reduced the original non-exact equation to a form where the solution can be expressed in terms of an integral.

Note: This solution might require numerical methods or specific techniques depending on the context and desired accuracy.