vestirme4

2021-02-24

$\left({x}^{2}+2xy-4{y}^{2}\right)dx-\left({x}^{2}-8xy-4{y}^{2}\right)dy=0$

grbavit

Since this is a homogeneous equation we put
$y=vx⇒dy=vdx+xdv$
Hence subtitiuting above we have
$\left({x}^{2}+2xy-4{y}^{2}\right)dx-\left({x}^{2}-8xy-4{y}^{2}\right)dy=0$
$⇒\left({x}^{2}+2v{x}^{2}-4{v}^{2}{x}^{2}\right)dx-\left({x}^{2}-8v{x}^{2}-4{v}^{2}{x}^{2}\right)\left(xdv+vdx\right)=0$
$⇒\left({x}^{2}+2v{x}^{2}-4{v}^{2}{x}^{2}\right)dx-\left({x}^{3}-8v{x}^{3}-4{v}^{2}{x}^{3}\right)dv-\left(v{x}^{2}-8{v}^{2}{x}^{2}-4{v}^{3}{x}^{2}\right)dx=0$
deviding both sides by ${x}^{2}$
$⇒\left(1+2v-4{v}^{2}\right)dx-\left(x-8vx-4{v}^{2}x\right)dv-\left(v-8{v}^{2}-4{v}^{3}\right)dx=0$
$⇒\left(1+2v-4{v}^{2}-v+8{v}^{2}+4{v}^{3}\right)dx-x\left(1-8v-4{v}^{2}\right)dv=0$
$⇒\left(1+v+4{v}^{2}+4{v}^{3}\right)dx=x\left(1-8v-4{v}^{2}\right)dv$
$⇒\frac{dx}{x}=\frac{1-8v-4{v}^{2}}{1+v+4{v}^{2}+4{v}^{3}}dv$
$⇒\frac{dx}{x}=\frac{1-8v-4{v}^{2}}{\left(1+v\right)\left(1+4{v}^{2}\right)}dv$
$⇒\frac{dx}{x}=\frac{\left(4{u}^{2}+1\right)-8u\left(u+1\right)}{\left(1+v\right)\left(1+4{v}^{2}\right)}$
$⇒\frac{dx}{x}=\left[\frac{4{u}^{2}+1}{\left(1+v\right)\left(1+4{v}^{2}\right)}-\frac{8u\left(u+1\right)}{\left(1+v\right)\left(1+4{v}^{2}\right)}\right]dv$
$⇒\frac{dx}{x}=\left[\frac{1}{\left(1+v\right)}-\frac{8u}{\left(1+4{v}^{2}\right)}\right]dv$

$⇒\mathrm{ln}\left(x\right)=\mathrm{ln}\left(1+v\right)-\mathrm{ln}\left(1+4{v}^{2}\right)+\mathrm{ln}C$
$⇒x=C\frac{1+v}{1+4{v}^{2}}$
C is the integrating constant
putting back the value of y
$⇒x=C\frac{1+\frac{y}{x}}{1+4\left(\frac{y}{x}{\right)}^{2}}$

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