Question

Non-exact equation and Integrating Factor

Given the equation below, solve for the general solutions.

w = dependent variable, v = independent variable

1. w(2v - w + 1) dv + v (3v - 4w + 3) dw = 0

Answer 1

$$P(v, w)=2vw-w^2+w$$

$$P_w=2v-2w+1$$

$$Q(v,w)=3v^2-4vw+3v$$

$$Q_v=6v-4w+3$$

$$P_w-Q_v=2v-2w+1-6v+4w-3$$

$$=-4v+2w-2$$

$$\dfrac{P_w-Q_v}{P}=\dfrac{-4v+2w-2}{2vw-w^2+w}=-\dfrac{2}{w}=-\psi(w)$$

$$\mu(w)=e^{\int{2 \over w}dw}=w^2$$

$$(2vw^3-w^4+w^3)dv+(3v^2w^2-4vw^3+3vw^2)dw=0$$

$$M_w=6vw^2-4w^3+3w^2$$

$$N_v=6vw^2-4w^3+3w^2$$

$$M_w=6vw^2-4w^3+3w^2=N_v$$

Then we write the system of two differential equations that define the function $u(v,w):$

$$u_v=M(v,w)$$

$$u_w=N(v,w)$$

$$u(v,w)=\int(2vw^3-w^4+w^3)dv+\varphi(w)$$

$$=v^2w^3-vw^4+vw^3+\varphi(w)$$

$$u_w=N(v,w)$$

$$u_w=3v^2w^2-4vw^3+3vw^2+\varphi'(w)$$

$$=3v^2w^2-4vw^3+3vw^2$$

Then

$$\varphi'(w)=0$$

$$\varphi(w)=C_1$$

The general solution of the differential equation

$$w(2v-w+1)dv+v(3v-4w+3)dw=0$$

is given by

$$v^2w^3-vw^4+vw^3=C$$