dx-%2B-e-x-y-(1-%E2%88%92-x-y-)dy-%3D-0.jpeg "(1 + E X Y )dx + E X Y (1 − X Y )dy = 0")
Solving the Differential Equation: (1 + e^xy)dx + e^xy(1 - xy)dy = 0
This article will guide you through solving the given differential equation:
(1 + e^xy)dx + e^xy(1 - xy)dy = 0
This equation is a first-order, non-linear differential equation. We can solve this using the method of exact differential equations.
1. Identifying an Exact Equation
A differential equation of the form:
M(x, y)dx + N(x, y)dy = 0
is considered exact if the following condition holds:
∂M/∂y = ∂N/∂x
In our case:
- M(x, y) = 1 + e^xy
- N(x, y) = e^xy(1 - xy)
Let's check the condition:
- ∂M/∂y = xe^xy
- ∂N/∂x = xe^xy
Since ∂M/∂y = ∂N/∂x, the given differential equation is exact.
2. Finding the Solution
For an exact differential equation, there exists 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) = ∫(1 + e^xy)dx = x + (e^xy)/y + f(y)
where f(y) is an arbitrary function of y.
Now, differentiating this expression with respect to y, we get:
∂u/∂y = xe^xy - (e^xy)/y^2 + f'(y)
Comparing this with N(x, y), we have:
xe^xy - (e^xy)/y^2 + f'(y) = e^xy(1 - xy)
Simplifying, we get:
f'(y) = -1/y
Integrating both sides with respect to y, we obtain:
f(y) = -ln|y| + C
where C is an arbitrary constant.
Substituting this value of f(y) back into the expression for u(x, y), we get the general solution:
u(x, y) = x + (e^xy)/y - ln|y| + C = 0
This is the implicit solution to the given differential equation.
Conclusion
We successfully solved the given differential equation by identifying it as an exact differential equation and utilizing the method for finding its solution. The solution, expressed implicitly, is:
x + (e^xy)/y - ln|y| + C = 0
This equation represents a family of curves that are the solutions to the original differential equation.