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Solving the Differential Equation (5x+4y)dx + (4x-8y^3)dy = 0
This article will guide you through the process of solving the given differential equation:
(5x+4y)dx + (4x-8y^3)dy = 0
This equation is a first-order, non-linear differential equation and can be solved using the method of exact differential equations.
1. Checking for Exactness
To determine if the equation is exact, we need to check if the following condition holds:
∂M/∂y = ∂N/∂x
Where:
- M(x, y) = 5x + 4y
- N(x, y) = 4x - 8y^3
Let's calculate the partial derivatives:
- ∂M/∂y = 4
- ∂N/∂x = 4
Since ∂M/∂y = ∂N/∂x, the given differential equation is exact.
2. Finding the Potential Function
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) = ∫(5x + 4y)dx = (5/2)x^2 + 4xy + g(y)
Where g(y) is an arbitrary function of y.
Now, differentiating this equation with respect to y and equating it to N(x, y), we get:
∂u/∂y = 4x + g'(y) = 4x - 8y^3
This implies that g'(y) = -8y^3. Integrating both sides, we get:
g(y) = -2y^4 + C
Where C is an arbitrary constant.
Substituting the value of g(y) back into the potential function, we get:
u(x, y) = (5/2)x^2 + 4xy - 2y^4 + C
3. The General Solution
The general solution of the exact differential equation is given by:
u(x, y) = C
Therefore, the general solution of the given differential equation is:
(5/2)x^2 + 4xy - 2y^4 = C
Where C is an arbitrary constant.
Conclusion
By using the method of exact differential equations, we have successfully solved the given differential equation:
(5x+4y)dx + (4x-8y^3)dy = 0
The general solution of this equation is:
(5/2)x^2 + 4xy - 2y^4 = C
Where C is an arbitrary constant.