Show weher the equation is exact and find its primitive
(3e^3x y - 2x)dx + e^3x dy=0
we want F such that
Fx = 3ye^(3x) - 2x
Fy = e^(3x)
Clearly F = ye^(3x)-x^2
To determine whether an equation is exact, we need to check if the partial derivatives of the terms involving x and y are equal. Let's start by rewriting the equation in a more standard form:
(3e^3x y - 2x)dx + e^3x dy = 0
Rearranging, we get:
(3e^3x y - 2x)dx = - e^3x dy
Taking the partial derivative of the term involving y with respect to x:
∂/∂y (3e^3x y - 2x) = 3e^3x
Note that the partial derivative of the term involving y is a function of x only, since y is not present. Now, taking the partial derivative of the term involving x with respect to y:
∂/∂x (- e^3x) = -3e^3x
Again, the partial derivative of the term involving x is solely a function of x.
Since the partial derivatives are equal (-3e^3x = 3e^3x), the equation is exact.
To find the primitive (or the general solution), we need to integrate either the term involving x or y with respect to their respective variables.
Let's integrate the term involving x:
∫(3e^3x y - 2x)dx = - ∫e^3x dy
Integrating the left side with respect to x:
∫(3e^3x y - 2x)dx = y∫3e^3x dx - ∫2x dx
= y*e^3x - x^2 + C1
Integrating the right side with respect to y:
- ∫e^3x dy = - y*e^3x + C2
Combining both sides:
y*e^3x - x^2 + C1 = - y*e^3x + C2
Simplifying, we have:
2y*e^3x - x^2 + C1 = C2
So the primitive (general solution) to the given equation is:
2y*e^3x - x^2 = C
where C is the constant of integration.