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

6 min read Jun 16, 2024

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

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

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

This equation is a first-order, non-linear ordinary differential equation. We can solve this by utilizing the concept of exact differential equations.

Understanding Exact Differential Equations

A differential equation of the form:

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

is considered exact if and only if:

∂M/∂y = ∂N/∂x

where ∂M/∂y represents the partial derivative of M with respect to y, and ∂N/∂x represents the partial derivative of N with respect to x.

Applying the Concept to Our Equation

Let's identify M(x, y) and N(x, y) from our given equation:

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

Now, let's calculate the partial derivatives:

∂M/∂y = 2y
∂N/∂x = 2xy

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

Finding an Integrating Factor

To make the equation exact, we need to find an integrating factor, which is a function μ(x, y) that, when multiplied by the original equation, will make it exact.

There are two common ways to find an integrating factor:

If (∂M/∂y - ∂N/∂x)/N is a function of x alone, then μ(x) = exp[∫((∂M/∂y - ∂N/∂x)/N)dx]
If (∂N/∂x - ∂M/∂y)/M is a function of y alone, then μ(y) = exp[∫((∂N/∂x - ∂M/∂y)/M)dy]

In our case:

((∂M/∂y - ∂N/∂x)/N) = (2y - 2xy) / (x^2y + y) = 2(1 - x) / (x^2 + 1) is a function of x alone.

Therefore, we can use the first formula to find the integrating factor:

μ(x) = exp[∫(2(1 - x) / (x^2 + 1))dx] = exp[2(arctan(x) - x/2)]

Solving the Exact Equation

Now, multiply the original equation by μ(x):

expdx + expdy = 0

This equation is now exact because:

∂(exp)/∂y = 2yexp[2(arctan(x) - x/2)]
∂(exp)/∂x = 2yexp[2(arctan(x) - x/2)]

Therefore, there exists a function F(x, y) such that:

∂F/∂x = exp
∂F/∂y = exp

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

F(x, y) = y^2 exp[2(arctan(x) - x/2)] + g(y)

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

Differentiating F(x, y) with respect to y and equating it to the second equation, we get:

2y exp

Solving for g'(y), we get:

g'(y) = y exp

Integrating g'(y) with respect to y, we get:

g(y) = (y^2/2) exp + C

where C is an arbitrary constant.

Therefore, the general solution of the differential equation is:

F(x, y) = y^2 exp + C = 0

This solution can be simplified and expressed in various forms depending on the desired level of detail or specific applications.