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Differential Equation: (y^4+2y)dx+(xy^3+2y^4-4x)dy=0
In this article, we will discuss the differential equation (y^4+2y)dx+(xy^3+2y^4-4x)dy=0. We will explore the characteristics of this equation and methods to solve it.
Form of the Equation
The given differential equation is of the form:
(M)y dx + (N)x dy = 0
where M = y^4+2y and N = xy^3+2y^4-4x.
Exact Differential Equation
A differential equation of the form (M)y dx + (N)x dy = 0 is said to be an exact differential equation if there exists a function u(x,y) such that:
∂u/∂x = M and ∂u/∂y = N
In this case, we can check if the given equation is exact by verifying the following condition:
∂M/∂y = ∂N/∂x
Solution to the Differential Equation
To solve the differential equation, we need to find the integrating factor μ(x,y) such that the equation becomes exact. Once we have the integrating factor, we can find the solution u(x,y).
After some analysis, we find that the integrating factor is μ(x,y) = 1/x. Multiplying both sides of the original equation by the integrating factor, we get:
(y^4/x+2y/x)dx+(xy^2+2y^4/x-4)dy=0
This equation is now exact, and we can find the solution u(x,y) by integrating both sides with respect to x and y.
Final Solution
After integrating and simplifying, we get the final solution:
u(x,y) = (y^4)/x + y^2 + C
where C is the constant of integration.
Conclusion
In this article, we have solved the differential equation (y^4+2y)dx+(xy^3+2y^4-4x)dy=0 by finding the integrating factor and then integrating both sides to obtain the final solution.
