Question

Resolver a equação diferencial ordinária (homogênea) usando método da substituição
HELP???

Answer 1

$\boxed{\begin{array}{l}\sf(3x^2-y^2)dy=2xydx\\\sf\dfrac{dy}{dx}=\dfrac{2xy}{3x^2-y^2}\\\sf\dfrac{dy}{dx}=\dfrac{2\backslash\!\!\!xy}{\backslash\!\!\!x^2\bigg(3-\bigg[\dfrac{y}{x}\bigg]^2\bigg)}\\\sf\dfrac{dy}{dx}=2\cdot\dfrac{y}{x}\cdot\dfrac{1}{3-\bigg[\dfrac{y}{x}\bigg]^2}\end{array}}$

$\boxed{\begin{array}{l}\underline{\rm fac_{\!\!,}a}\\\sf v=\dfrac{y}{x}\implies y=vx\\\sf\dfrac{dy}{dx}=\dfrac{d}{dx}(v\cdot x)\\\sf=\dfrac{dv}{dx}\cdot x+v\cdot\dfrac{dx}{dx}\\\sf\dfrac{d}{dx}(v\cdot x)=x\dfrac{dv}{dx}+v\\\underline{\rm substituindo~na~equac_{\!\!,}\tilde ao~diferencial~temos:}\end{array}}$

$\boxed{\begin{array}{l}\sf x\dfrac{dv}{dx}+v=2v\cdot\dfrac{1}{3-v^2}\\\sf x\dfrac{dv}{dx}=\dfrac{2v}{3-v^2}-v\\\sf x\dfrac{dv}{dx}=\dfrac{2v-3v+v^3}{3-v^2}\\\sf x\dfrac{dv}{dx}=\dfrac{v^3-v}{3-v^2}\\\sf\dfrac{3-v^2}{v^3-v}dv=\dfrac{dx}{x}\end{array}}$

$\boxed{\begin{array}{l}\sf\dfrac{3-v^2}{v^3-v}=\dfrac{3-v^2}{v\cdot(v^2-1)}=\dfrac{3-v^2}{v\cdot(v-1)\cdot (v+1)}\\\sf \dfrac{3-v^2}{v\cdot(v-1)\cdot(v+1)}=\dfrac{A}{v}+\dfrac{B}{v-1}+\dfrac{C}{v+1}\\\sf A=\dfrac{3-v^2}{(v-1)(v+1)}\bigg|_{v=0} =\dfrac{3}{(0-1)(0+1)}=\dfrac{3}{-1}=-3\\\\\sf B=\dfrac{3-v^2}{v\cdot(v+1)}\bigg |_{v=1}=\dfrac{3-1^2}{1\cdot(1+1)}=\dfrac{2}{2}=1\\\\\sf C=\dfrac{3-v^2}{v\cdot(v-1)}\bigg|_{v=-1}=\dfrac{3-(-1)^2}{-1\cdot(-1-1)}=\dfrac{2}{2}=1\end{array}}$

$\boxed{\begin{array}{l}\sf\dfrac{3-v^2}{v^3-v}=-\dfrac{3}{v}+\dfrac{1}{v-1}+\dfrac{1}{v+1}\\\underline{\rm substituindo~na~equac_{\!\!,}\tilde ao~diferencial~temos:}\end{array}}$

$\boxed{\begin{array}{l}\sf\bigg(-\dfrac{3}{v}+\dfrac{1}{v-1}+\dfrac{1}{v+1}\bigg)dv=\dfrac{dx}{x}\\\underline{\rm integrando~em~ambos~lados~temos:}\\\displaystyle\sf\int\bigg(-\dfrac{3}{v}+\dfrac{1}{v-1}+\dfrac{1}{v+1}\bigg)dv=\int\dfrac{dx}{x}\\\displaystyle\sf-3\int\dfrac{dv}{v}+\int\dfrac{dv}{v-1}+\int\dfrac{dv}{v+1}=\int\dfrac{dx}{x}\\\sf -3\ell n|v|+\ell n|v-1|+\ell n|v+1|=\ell n|x|+c\end{array}}$

$\boxed{\begin{array}{l}\sf \ell n\bigg|\dfrac{(v-1)\cdot(v+1)}{v^3}\bigg|=\ell n|x|+c\\\sf \bf{e}^{\sf\ell n|\frac{v^2-1}{v^3}|}=\bf{e}^{\sf\ell n|x|+c}\\\sf\dfrac{v^2-1}{v^3}=k\cdot x\\\sf \dfrac{(\frac{y}{x})^2-1}{(\frac{y}{x})^3}=k\cdot x\\\sf\dfrac{\frac{y^2}{x^2}-1}{\frac{y^3}{x^3}}=kx\\\sf\dfrac{\frac{y^2-x^2}{x^2}}{\frac{y^3}{x^3}}=kx\\\sf\dfrac{(y^2-x^2)}{\backslash\!\!\!x^2}\cdot\dfrac{\backslash\!\!\!x^3}{y^3}=kx\\\sf\dfrac{(y^2-x^2)\cdot\backslash\!\!\!x}{y^3}=k\cdot\backslash\!\!\!x\end{array}}$

$\blue{\large\boxed{\begin{array}{l}\sf\dfrac{y^2-x^2}{y^3}=k\\\sf S=\bigg\{ \dfrac{y^2-x^2}{y^3}=k\bigg\}\end{array}}}$