dx-2xydy%3D0.jpeg "(x^2+3y^2)dx-2xydy=0")
Differential Equation: (x^2+3y^2)dx-2xydy=0
Introduction
In this article, we will discuss the differential equation (x^2+3y^2)dx-2xydy=0. This is a type of first-order ordinary differential equation, which is widely used in various fields such as physics, engineering, and mathematics.
The Equation
The differential equation is given by:
(x^2+3y^2)dx-2xydy=0
This equation can be written in the form:
Mdx + Ndy = 0
where M = x^2+3y^2 and N = -2xy.
Exact Differential Equation
A differential equation of the form Mdx + Ndy = 0 is said to be exact if there exists a function f(x,y) such that:
∂f/∂x = M and ∂f/∂y = N
In this case, we can show that the differential equation is exact by finding the function f(x,y) that satisfies the above conditions.
Solving the Equation
To solve the differential equation, we can use the method of exact differential equations. First, we find the integral of M with respect to x, which is:
∫Mdx = ∫(x^2+3y^2)dx = (x^3/3 + x^2y^2) + C1
where C1 is a constant of integration.
Next, we find the integral of N with respect to y, which is:
∫Ndy = ∫(-2xy)dy = -x^2y + C2
where C2 is a constant of integration.
Now, we equate the two expressions and solve for y:
(x^3/3 + x^2y^2) + C1 = -x^2y + C2
Simplifying the equation, we get:
x^2y^2 + x^2y = C
where C = C2 - C1 is a constant.
Conclusion
In this article, we have discussed the differential equation (x^2+3y^2)dx-2xydy=0 and shown that it is an exact differential equation. We have also solved the equation using the method of exact differential equations. The general solution of the equation is given by x^2y^2 + x^2y = C, where C is a constant.
