Question

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

Answer 1

$\boxed{\begin{array}{l}\sf(4x-3y)dx+(2y-3x)dy=0\\\sf (2y-3x)dy=(3x-4y)dx\\\sf\dfrac{dy}{dx}=\dfrac{3x-4y}{2y-3x}\\\sf\dfrac{dy}{dx}=\dfrac{\backslash\!\!\!x\cdot(3-4\cdot\frac{y}{x})}{\backslash\!\!\!x(2\cdot\frac{y}{x}-3)}\\\sf\dfrac{dy}{dx}=\dfrac{3-4\cdot\frac{y}{x}}{(2\cdot\frac{y}{x}-3)}\end{array}}$

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

$\boxed{\begin{array}{l}\sf x\dfrac{dv}{dx}+v=\dfrac{3-4v}{2v-3}\\\sf x\dfrac{dv}{dx}=\dfrac{3-4v}{2v-3}-v\\\sf x\dfrac{dv}{dx}=\dfrac{3-4v-2v^2+3v}{2v-3}\\\sf x\dfrac{dv}{dx}=\dfrac{-2v^2-v+3}{2v-3}\\\sf\dfrac{2v-3}{-2v^2-v+3}dv=\dfrac{dx}{x}\end{array}}$

$\boxed{\begin{array}{l}\sf\dfrac{2v-3}{-2v^2-v+3}=\dfrac{2v-3}{(v-1)\cdot(-2v-3)}\\\sf\dfrac{2v-3}{-2v^2-v+3}=\dfrac{A}{v-1}+\dfrac{B}{-2v-3}\\\sf2v-3=A(-2v-3)+B(v-1)\\\sf 2v-3=-2Av-3A+Bv-B\\\sf 2v-3=(-2A+B)v-3A-B\\+\underline{\begin{cases}\sf-2A+B=2\\\sf -3A-B=-3\end{cases}}\\\sf-5A=-1\cdot(-1)\\\sf A=\dfrac{1}{5}\\\sf -2A+B=2\\\sf-2\cdot\dfrac{1}{5}+B=2\cdot5\\\sf -2+5B=10\\\sf 5B=10+2\\\sf 5B=12\\\sf B=\dfrac{12}{5}\end{array}}$

$\boxed{\begin{array}{l}\sf\dfrac{2v-3}{(v-1)(-2v-3)}=\dfrac{1}{5}\cdot\dfrac{1}{v-1}+\dfrac{12}{5}\cdot\dfrac{1}{-2v-3}\\\underline{\rm substituindo~na~equac_{\!\!,}\tilde ao~diferencial~temos:}\end{array}}$

$\boxed{\begin{array}{l}\sf\bigg(\dfrac{1}{5}\cdot\dfrac{1}{v-1}+\dfrac{12}{5}\cdot\dfrac{1}{-2v-3}\bigg)dv=\dfrac{dx}{x}\\\underline{\rm integrando~dos~dois~lados~temos\!:}\\\displaystyle\sf\int\bigg(\dfrac{1}{5}\cdot\dfrac{1}{v-1}+\dfrac{12}{5}\cdot\dfrac{1}{-2v-3}\bigg)dv=\int\dfrac{dx}{x}\\\displaystyle\sf\dfrac{1}{5}\int\dfrac{dv}{v-1}+\dfrac{12}{5}\int\dfrac{dv}{-2v-3}=\int\dfrac{dx}{x}\end{array}}$

$\boxed{\begin{array}{l}\displaystyle\sf\dfrac{1}{5}\int\dfrac{dv}{v-1}\\\rm fac_{\!\!,}a\\\sf t=v-1\implies dv=dt\\\displaystyle\sf\dfrac{1}{5}\int\dfrac{dv}{v-1}=\dfrac{1}{5}\int\dfrac{dt}{t}=\dfrac{1}{5}\ell n|t|+c\\\displaystyle\sf\dfrac{1}{5}\int\dfrac{dv}{v-1}=\dfrac{1}{5}\ell n|v-1|+c\end{array}}$

$\boxed{\begin{array}{l}\displaystyle\sf\dfrac{12}{5}\int\dfrac{dv}{-2v-3}=\dfrac{\diagdown\!\!\!\!\!\!12}{5}\cdot-\dfrac{1}{\diagdown\!\!\!\!2}\int\dfrac{-2dv}{-2v-3}\\\rm fac_{\!\!,}a\\\sf t=-2v-3\implies -2dv=dt\\\displaystyle\sf-\dfrac{6}{5}\int\dfrac{-2dv}{-2v-3}=-\dfrac{6}{5}\int\dfrac{dt}{t}=-\dfrac{6}{5}\ell n|t|+c\\\displaystyle\sf\dfrac{12}{5}\int\dfrac{dv}{-2v-3}=-\dfrac{6}{5}\ell n|-2v-3|+c\end{array}}$

$\boxed{\begin{array}{l}\underline{\rm substituindo~os~resultados~}\\\underline{\rm na~equac_{\!\!,}\tilde ao~diferencial~temos:}\\\sf\dfrac{1}{5}\ell n|v-1|-\dfrac{6}{5}\ell n|-2v-3|=\ell n|x|+c\\\sf\ell n\bigg|\dfrac{(v-1)^{\frac{1}{5}}}{(-2v-3)^{\frac{6}{5}}}\bigg|=\ell n|x|+k\\\bf{e}^{\sf\ell n\bigg|\frac{(v-1)^{\frac{1}{5}}}{(-2v-3)^{\frac{6}{5}}}\bigg|}=\bf{e}^{\sf\ell n|x|+k}\end{array}}$

$\boxed{\begin{array}{l}\sf\dfrac{(v-1)^{\frac{1}{5}}}{(-2v-3)^{\frac{6}{5}}}=kx\\\sf\dfrac{(\frac{y}{x}-1)^{\frac{1}{5}}}{(-2\cdot\dfrac{y}{x}-3)^{\frac{6}{5}}}=kx\\\sf\dfrac{(\frac{y-x}{x})^{\frac{1}{5}}}{(\frac{-2y-3x}{x})^{\frac{6}{5}}}=kx\end{array}}$