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Solving the Differential Equation: (2xy² - 3)dx + (2x²y + 4)dy = 0
This article will guide you through solving the given differential equation:
(2xy² - 3)dx + (2x²y + 4)dy = 0
This equation is classified as an exact differential equation. To understand this, let's recall some definitions:
Exact Differential Equation: A differential equation of the form:
M(x, y)dx + N(x, y)dy = 0
is said to be exact if:
∂M/∂y = ∂N/∂x
where ∂M/∂y represents the partial derivative of M with respect to y, and ∂N/∂x represents the partial derivative of N with respect to x.
Let's check if our given equation is exact:
- M(x, y) = 2xy² - 3
- N(x, y) = 2x²y + 4
Calculating the partial derivatives:
- ∂M/∂y = 4xy
- ∂N/∂x = 4xy
Since ∂M/∂y = ∂N/∂x, the equation is exact.
Solving the Exact Equation:
- Find a function F(x, y) such that:
∂F/∂x = M(x, y) and ∂F/∂y = N(x, y)
- Integrate ∂F/∂x = M(x, y) with respect to x, treating y as a constant:
F(x, y) = ∫(2xy² - 3) dx = x²y² - 3x + g(y)
Where g(y) is an arbitrary function of y.
- Differentiate F(x, y) with respect to y and equate it to N(x, y):
∂F/∂y = 2x²y + g'(y) = 2x²y + 4
- Solve for g'(y):
g'(y) = 4
- Integrate g'(y) with respect to y to find g(y):
g(y) = ∫4 dy = 4y + C
where C is an arbitrary constant.
- Substitute g(y) back into the expression for F(x, y):
F(x, y) = x²y² - 3x + 4y + C
The general solution to the exact differential equation is given by:
F(x, y) = C
Therefore, the solution to the given differential equation (2xy² - 3)dx + (2x²y + 4)dy = 0 is:
x²y² - 3x + 4y = C
where C is an arbitrary constant.