dx-xdy%3D0.jpeg "(x^4+2y)dx-xdy=0")
Differential Equation: (x^4+2y)dx-xdy=0
Introduction
In this article, we will discuss the solution to the differential equation (x^4+2y)dx-xdy=0. Differential equations are a fundamental concept in mathematics and are used to model various phenomena in fields such as physics, engineering, and economics.
The Differential Equation
The differential equation we are interested in is:
(x^4+2y)dx-xdy=0
This is a first-order ordinary differential equation, where x and y are the variables, and dx and dy are their respective differentials.
Solution to the Differential Equation
To solve this differential equation, we can use various methods such as separation of variables, integration, or substitution. Here, we will use the substitution method to find the general solution.
Let's assume y = vx, where v is a function of x. Then, we have:
dy = vdx + xdv
Substituting this into the original differential equation, we get:
(x^4+2vx)dx-x(vdx + xdv) = 0
Simplifying the equation, we get:
x^4dx + 2vx^2dx - vx^2dx - x^2dv = 0
Combining like terms, we get:
x^4dx + vx^2dx - x^2dv = 0
Dividing both sides by x^2, we get:
x^2dx + vdx - dv = 0
This is a separable differential equation, and we can separate the variables as follows:
∫(x^2 + v)dx = ∫dv
Evaluating the integral, we get:
(x^3/3 + vx) = v + C
where C is the constant of integration.
Solving for v, we get:
v = (x^3/3 + C)/(x - 1)
Substituting v = y/x, we get the general solution:
y = (x^3/3 + C)/(x - 1)
Conclusion
In this article, we have solved the differential equation (x^4+2y)dx-xdy=0 using the substitution method. The general solution is y = (x^3/3 + C)/(x - 1), where C is the constant of integration. This solution can be used to model various phenomena in fields such as physics and engineering.
