Answer to Question #209129 in Differential Equations for jay

Question

w=dependent variable

v=independendt variable

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

Answer 1

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

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

$M(v,w)=(2vw-w^2+w) \implies {\partial M(v,w) \over \partial w}=2v-2w+1$

$N(v,w)=(3v^2-4vw+3v) \implies {\partial N(v,w) \over \partial v}=6v-4w+3$

The differential equation is not exact since ${\partial M(v,w) \over \partial w} \not = {\partial N(v,w) \over \partial v}$

There is no integrating factor of the form $p(v)$ or $p(w)$

Let's try another method

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

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

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

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

$2vw-w^2+w+3v^2{dw\over dv}-4vw{dw\over dv}+3v{dw\over dv}=0 ..........(i)$

Let $u=vw\implies {du\over dv}=w+v{dw\over dv} \implies {dw\over dv}={{du\over dv}\over v}-{u\over v^2}$ and Substituting this to equation $(i)$

$2v{u\over v}-({u\over v})^2+({u\over v})+3v^2({{du\over dv}\over v}-{u\over v^2})-4v{u\over v}({{du\over dv}\over v}-{u\over v^2}) +3v({{du\over dv}\over v}-{u\over v^2})=0$

The Variables cannot be separated further

Lets check if the differential equation is homogenous

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

$(3v^2-4vw+3v){dw\over dv}=-(2vw-w^2+w)$

${dw\over dv}=-{(2vw-w^2+w)\over (3v^2-4vw+3v)}$

The differential equation is not homogenous

$\therefore$ The differential equation has no solution