Determine whether the equation is exact If it is exact find

Determine whether the equation is exact. If it is exact, find the solution. (2x^2-2xy+7)+(5y^2-x^2+2)y\'=0

Solution

(2x²-2xy+7)+(5y²-x²+2)y\'=0

(2x²-2xy+7)dx+(5y²-x²+2)dy=0

f_x=M=2x²-2xy+7

M_y=0-2x+0

M_y=-2x

f_y=N=5y²-x²+2

N_x=0-2x+0

N_x=-2x

M_y=N_x

so equation is exact

solution is of form

f= 2x²-2xy+7 dx

f=(2/3)x³-x²y+7x +g(y)

differentiate partially with respect to y

f_y=0-x²+0+g_y(y)

we have f_y=N=5y²-x²+2

0-x²+0+g_y(y)=5y²-x²+2

g_y(y)=5y²+2

g(y)= 5y²+2 dy

g(y)= (5/3)y³+2y +c

f=(2/3)x³-x²y+7x +g(y)

f=(2/3)x³-x²y+7x + (5/3)y³+2y +c

(2/3)x³-x²y+7x + (5/3)y³+2y =c is the solution