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

4 min read Jun 17, 2024

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

This article will walk you through the process of solving the first-order differential equation:

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

1. Recognizing the Type of Differential Equation

The given equation is a non-exact differential equation. This means that it doesn't satisfy the condition for exactness:

∂M/∂y = ∂N/∂x

Where:

M(x,y) = x-y-2
N(x,y) = -(2x-2y-3)

In our case, ∂M/∂y = -1 and ∂N/∂x = -2, which are not equal.

2. Finding an Integrating Factor

To solve this non-exact differential equation, we need to find an integrating factor, which will make the equation exact.

a) Finding the Integrating Factor (IF):

We can use the following formula to find the integrating factor:

IF = exp(∫(∂N/∂x - ∂M/∂y) / M dx)

In our case, this becomes:

IF = exp(∫((-2) - (-1)) / (x-y-2) dx)
IF = exp(∫(-1) / (x-y-2) dx)
IF = exp(-ln|x-y-2|)
IF = 1/(x-y-2)

b) Multiplying the Equation by the Integrating Factor:

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

(1/(x-y-2)) * [(x-y-2)dx - (2x-2y-3)dy] = 0
dx - [(2x-2y-3)/(x-y-2)]dy = 0

This equation is now exact.

3. Solving the Exact Differential Equation

Let's rewrite the equation as:

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

Where:

M(x,y) = 1
N(x,y) = -(2x-2y-3)/(x-y-2)

Now, we need to find a function u(x,y) such that:

∂u/∂x = M(x,y)
∂u/∂y = N(x,y)

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

u(x,y) = x + g(y)

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

Differentiating this expression with respect to y and equating it to N(x,y), we get:

∂u/∂y = g'(y) = -(2x-2y-3)/(x-y-2)

This equation doesn't contain x, so we can solve it for g'(y) by setting x = 0:

g'(y) = (2y+3)/(y+2)

Integrating g'(y) with respect to y:

g(y) = 2y + ln|y+2| + C

Finally, substituting this expression for g(y) back into the equation for u(x,y), we get the solution:

u(x,y) = x + 2y + ln|y+2| + C = 0

4. Implicit Solution

The solution to the original differential equation is given implicitly by the equation:

x + 2y + ln|y+2| + C = 0

Where C is an arbitrary constant.

Conclusion

We successfully solved the non-exact differential equation (x-y-2)dx-(2x-2y-3)dy=0 by finding an integrating factor, making the equation exact, and then integrating to obtain the implicit solution.