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Solving the Differential Equation: (2x^2y-3y^4)dx+(3x3+2xy3)dy=0
In this article, we will discuss how to solve the differential equation (2x^2y-3y^4)dx+(3x3+2xy3)dy=0.
What is a Differential Equation?
A differential equation is a mathematical equation that involves an unknown function and its derivatives. It is a fundamental concept in mathematics and is used to model various phenomena in physics, engineering, and other fields.
The Given Differential Equation
The given differential equation is:
(2x^2y-3y^4)dx+(3x3+2xy3)dy=0
This is a first-order partial differential equation, where x and y are the independent variables.
Solving the Differential Equation
To solve this differential equation, we can use various methods, such as the method of separation of variables or the method of characteristics. Here, we will use the method of separation of variables.
Step 1:Separate the Variables
Rearrange the equation to separate the variables:
(2x^2y-3y^4)dx = -(3x3+2xy3)dy
Step 2:Integrate Both Sides
Integrate both sides of the equation with respect to x and y, respectively:
∫(2x^2y-3y^4)dx = -∫(3x3+2xy3)dy
Step 3:Solve for the Constant
Solve for the constant of integration:
x^2y - y^4/4 = -x^4/4 - xy^3/3 + C
where C is the constant of integration.
Conclusion
In this article, we have solved the differential equation (2x^2y-3y^4)dx+(3x3+2xy3)dy=0 using the method of separation of variables. The general solution is given by x^2y - y^4/4 = -x^4/4 - xy^3/3 + C, where C is the constant of integration.
