Solving First Order Non-Exact Differential Equations

Understanding and solving first-order differential equations is a fundamental skill in mathematics, particularly in an applied context. This article illustrates a method for solving non-exact differential equations through the use of integrating factors. Specifically, we will explore a detailed solution to the differential equation Mxydy - Nxydx 0, where M y/x and N y^2 - x^2. The process involves checking if the equation is exact, determining the appropriate integrating factor, and then solving the resulting exact differential equation.

Initial Setup

The given differential equation is:

Mxydy - Nxydx 0

With M y/x and N y^2 - x^2, we can start by finding Mx and Ny:

Mx -y/x^2

Ny 2y

Since Ny - Mx/M 2x^2 - 1/x, which depends only on x, this helps us to formulate an integrating factor. The integrating factor (IF) is given by:

IF exp int; (2x^2 - 1/x) dx

Calculating the Integrating Factor

In order to find the integrating factor, we need to integrate 2x^2 - 1/x with respect to x:

int; (2x^2 - 1/x) dx (2/3)x^3 - ln|x| c

Therefore, the integrating factor is:

IF exp ((2/3)x^3 - ln|x| c) (x^2/e^(2/3)x^3) * exp(c)

Since the constant can be absorbed into a new constant, we can simplify this to:

IF x exp(x^2)

Transforming the Equation

With the integrating factor x exp(x^2), the new differential equation becomes:

(y exp(x^2) dx x exp(x^2)dy) 0

We now need to check if this new equation is exact. To do this, we find P y exp(x^2) and Q x exp(x^2) - 1 and check Px Qy:

Px y * 2x exp(x^2) 2xy exp(x^2)

Qy x * 2x exp(x^2) 2x^2 exp(x^2)

Since Px Qy, the equation is now exact. The solution to this equation can be found by integrating (y exp(x^2) dx x exp(x^2)dy) 0. We let:

F(x, y) C

where dF/dx y exp(x^2) and dF/dy x exp(x^2).

Solving the Exact Equation

Introducing the potential function F(x, y), we integrate each term with respect to the corresponding variable:

Fy y * exp(x^2)

Fx x * exp(y^2 - x^2 - 1)

To integrate, we consider Fy y exp(x^2):

int; y exp(x^2) dy Fxy g(x) y^2/2 exp(x^2) g(x)

Next, we use the other equation Fx to determine g(x):

g'(x) -x * exp(x^2) 1/2 * x int; exp(x^2) dx constant

The integral simplifies to:

g(x) (1/2 - exp(x^2)) * x^2 constant

Therefore, the solution to the differential equation is:

x^2 y^2/2 - exp(x^2) C

Second Example: Solving a Different First Order Non-Exact Equation

Next, we consider the differential equation:

ty' - (1 x^2)y 1 - x^2

Setting x^2 y^2 t and y' 1 - (1 x^2) / y, we derive:

dy/dx (1 - t) y / x^2 y

By transforming, we find:

1/2 - t dt 2x dx

Integrating both sides:

-log(2 - t) x^2 C

Since t x^2 y^2, we get:

-log(2 - x^2 y^2) x^2 C

Hence the solution is:

log(2 - x^2 y^2) -x^2 C