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Differential Equation: (cos2y-3x^2y^2)dx+(cos2y-2x sin 2y-2x^3y)dy=0
In this article, we will discuss the differential equation (cos2y-3x^2y^2)dx+(cos2y-2x sin 2y-2x^3y)dy=0 and its solution.
Introduction
A differential equation is a mathematical equation that involves an unknown function and its derivatives. It is a fundamental concept in calculus and is used to model various phenomena in fields such as physics, engineering, and economics. In this article, we will focus on the differential equation (cos2y-3x^2y^2)dx+(cos2y-2x sin 2y-2x^3y)dy=0.
Form of the Differential Equation
The given differential equation is of the form:
M(x,y)dx + N(x,y)dy = 0
where M(x,y) = cos2y-3x^2y^2 and N(x,y) = cos2y-2x sin 2y-2x^3y.
Solution of the Differential Equation
To solve this differential equation, we can use the method of separation of variables. This method involves rearranging the equation to separate the variables x and y.
First, we can rewrite the equation as:
(cos2y-3x^2y^2)dx = -(cos2y-2x sin 2y-2x^3y)dy
Next, we can divide both sides of the equation by (cos2y-3x^2y^2):
dx = -[(cos2y-2x sin 2y-2x^3y)/(cos2y-3x^2y^2)]dy
Now, we can integrate both sides of the equation with respect to x and y, respectively:
∫dx = -∫[(cos2y-2x sin 2y-2x^3y)/(cos2y-3x^2y^2)]dy
x + C1 = -∫[(cos2y-2x sin 2y-2x^3y)/(cos2y-3x^2y^2)]dy + C2
where C1 and C2 are constants of integration.
Conclusion
In this article, we have discussed the differential equation (cos2y-3x^2y^2)dx+(cos2y-2x sin 2y-2x^3y)dy=0 and its solution using the method of separation of variables. The solution involves integrating both sides of the equation with respect to x and y, respectively, and rearranging the equation to obtain the general solution.
