Find the value of b for which ye2xy x bxe2xy y 0 is exact

Find the value of b for which (ye^2xy + x) + bxe^2xy y\' = 0 is exact, and then find a solution using the determined value for b.

Solution

Equation is of the form

Mdx+Ndy=0

M=y exp(2xy)

N=b x exp(2xy)

For exact ness we need

M_y=N_x

M_y=2xy exp(2xy)+exp(2xy)

N_x=b exp(2xy)+2bxy exp(2xy)

So for equation to be exact

b=1

So integral is

F(x,y)=C

So that

dF=Mdx+Ndy=0

F_x=M=y exp(2xy)+x

Integrating w.r.t. x treating y as a parameter.

F= exp(2xy)/2+x^2/2+g(y)

So,

F_y=x exp(2xy)+g\'=N

F_y=N=x exp(2xy)

Hence, g\'=0 is g is constant so

F= exp(2xy)/2+x^2/2=C