Verify that the equation is exact and then solve the equatio

Verify that the equation is exact and then solve the equation (2x + y^2)dx + (2xy)dy = 0.

Solution

In order to prove that a differential equation is exact, we have to have the partial derivatives equal to each other.

(2x + y^2)dx + (2xy)dy = 0

Our equation is of form M(x,y) dx + N(x,y) dy = 0, M and N being functions of x and y

M(x,y) = 2x + y^2   and N(x,y) = 2xy

In order to check if the equation is exact, we need to show that the partial derivative of M with respect to y must be equal to the partial derivative of N with respect to x.

My = 2y   and Nx = 2y

SInce M(x,y)_y =N(x,y)_x

hence the given equation is exact differential equation.

the general solution f(x,y) = C is given by

f(x,y) = integral [M(x,y)]dx = integral [2x + y^2]dx = x^2 + xy^2+ g(y)

now f_y(x,y) = 2xy + g\'(y)

=> g\'(y) = 0

=> g(y) = C , C is a constant

=> f(x,y) = x^2 + xy^2+ C

and the general solution is x^2 + xy^2 = -C