Author Topic: TUT0202 Quiz 1  (Read 866 times)

Jiwen Bi

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TUT0202 Quiz 1
« on: September 27, 2019, 03:35:10 PM »
${y}'=\frac{x^{2}+3y^{2}}{2xy}\\=\frac{x}{2y}+\frac{3y}{2x}=\frac{1}{2}(\frac{y}{x})^{-1}+\frac{3}{2}(\frac{y}{x})$

let  $v=\frac{y}{x}\\{v}'x+v={y}'\\
{v}'x+v=\frac{1}{2}v^{-1}+\frac{3}{2}v\\
{v}'x=\frac{1}{2}v^{-1}+\frac{1}{2}v=\frac{1+v^{2}}{2v}$
Equation is homogeneous
$\frac{2v}{1+v^{2}}dv=\frac{1}{x}\\
ln\left | v^{2}+1 \right |=ln\left | x \right |+c\\
v^{2}+1=cx\\
\frac{y^{2}}{x^{2}}+1-cx=0\\
y^{2}+x^{2}-cx^{3}=0$
« Last Edit: September 27, 2019, 07:59:51 PM by Jiwen Bi »