Answer to Question #209482 in Differential Equations for Karylle

Question

 w (4v + w) dv - 2 (v^2 - w) dw = 0

Answer 1

$$P(v, w)=4vw+w^2$$

$$P_w=4v+2w$$

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

$$Q_v=-4v$$

$$P_w-Q_v=4v+2w+4v=8v+2w$$

$$\dfrac{P_w-Q_v}{P}=\dfrac{8v+2w}{4vw+w^2}=\dfrac{2}{w}=-\psi(w)$$

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

$$(4\dfrac{v}{w}+1)dv+(-2\dfrac{v^2}{w^2}+2\dfrac{1}{w})dw=0$$

$$M_w=-4\dfrac{v}{w^2}$$

$$N_v=-4\dfrac{v}{w^2}$$

$$M_w=-4\dfrac{v}{w^2}=N_v$$

$$u(v,w)=\int(4\dfrac{v}{w}+1)dv+\varphi(w)$$

$$=2\dfrac{v^2}{w}+v+\varphi(w)$$

$$u_w=-2\dfrac{v^2}{w^2}+\varphi'(w)$$

$$=-2\dfrac{v^2}{w^2}+2\dfrac{1}{w}$$

$$\varphi'(w)=2\dfrac{1}{w}$$

$$\varphi(w)=2\ln w+C_1$$

The general solution of the differential equation

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

is given by

$$2\dfrac{v^2}{w}+v+\ln w^2=C$$