Find the integrating factor that is a function of x or y alone and use it to find the general solution of the differential equation?

You have differential equation: M(x,y) dx + N(x,y) dy = 0

Equation is exact if ∂M/∂y = ∂N/∂x

M(x,y) = x + y ----> ∂M/∂y = 1

N(x,y) = tan(x) ----> ∂N/∂x = sec²x

Equation is not exact

We can find integrating factor for these 2 cases:

• (∂M/∂y − ∂N/∂x) / N is a function of x only

　Integrating factor = e^(∫ (∂M/∂y−∂N/∂x)/N dx)

• (∂N/∂x − ∂M/∂y) / M is a function of y only

　Integrating factor = e^(∫ (∂N/∂x−∂M/∂y)/M dy)

(∂M/∂y − ∂N/∂x) / N = (1−sec²x)/tan(x) = −tan(x)

e^(∫ −tan(x) dx) = e^(ln cos(x)) = cosx

Multiply differential equation by cosx:

(x cosx + y cosx) dx + sinx dy = 0

M(x,y) = x cosx + y cosx ----> ∂M/∂y = cos x

N(x,y) = sin x ----> ∂N/∂x = cos x

Equation is now exact

Solution: F(x,y) = C, where

∂F/∂x = M(x,y) = x cosx + y cosx

∂F/∂y = N(x,y) = sinx

Solution: x sinx + cosx + y sinx = C