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

Suzhoumeite

4 min read

·

Jun 07, 2024

35

Share

Exact Differential Equations: Solving (1-xy)^-2 dx + (y^2+x^2(1-xy)^-2)dy=0

Introduction

In this article, we will explore the solution of the differential equation (1-xy)^-2 dx + (y^2+x^2(1-xy)^-2)dy=0, which is a type of exact differential equation. We will discuss the concept of exact differential equations and how to solve them using the method of integrating factors.

What is an Exact Differential Equation?

A differential equation of the form M(x,y)dx + N(x,y)dy = 0 is said to be exact if there exists a function ψ(x,y) such that ∂ψ/∂x = M(x,y) and ∂ψ/∂y = N(x,y). In other words, an exact differential equation is one that can be expressed as the total differential of a function.

Testing for Exactness

Before we can solve the differential equation, we need to test whether it is exact. To do this, we need to check if the following condition is satisfied:

∂M/∂y = ∂N/∂x

In our case, we have:

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

Computing the partial derivatives, we get:

∂M/∂y = 2xy(1-xy)^-3 and ∂N/∂x = -2y(1-xy)^-3 + 2x(1-xy)^-2

Substituting the values, we get:

∂M/∂y = ∂N/∂x

This shows that the differential equation is exact.

Solving the Differential Equation

Since the differential equation is exact, we can write it in the form:

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

To solve this equation, we need to find the function ψ(x,y) such that ∂ψ/∂x = (1-xy)^-2 and ∂ψ/∂y = y^2+x^2(1-xy)^-2.

Integrating ∂ψ/∂x = (1-xy)^-2 with respect to x, we get:

ψ(x,y) = ∫(1-xy)^-2 dx + f(y)

Using the substitution u = 1-xy, we get:

ψ(x,y) = -ln|1-xy| + f(y)

Now, differentiating ψ(x,y) with respect to y, we get:

∂ψ/∂y = -x/(1-xy) + f'(y)

Comparing this with ∂ψ/∂y = y^2+x^2(1-xy)^-2, we get:

f'(y) = y^2 and f(y) = (1/3)y^3 + C

Substituting f(y) into ψ(x,y), we get:

ψ(x,y) = -ln|1-xy| + (1/3)y^3 + C

Conclusion

In this article, we have solved the differential equation (1-xy)^-2 dx + (y^2+x^2(1-xy)^-2)dy=0 using the method of integrating factors. We have shown that the differential equation is exact and found the general solution ψ(x,y) = -ln|1-xy| + (1/3)y^3 + C.