Question

Resolver as ED homogênea
Matéria: equação diferencial e integral do curso de eng
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Answer 1

$\boxed{\begin{array}{l}\rm 1)~\sf xdx+(y-2x)dy=0\\\sf(y-2x)dy=-xdx\\\sf\dfrac{dy}{dx}=-\dfrac{x}{y-2x}\\\sf\dfrac{dy}{dx}=\dfrac{x}{2x-y}\\\sf\dfrac{dy}{dx}=\dfrac{\backslash\!\!\!x}{\backslash\!\!\!x(2-\frac{y}{x})}\\\sf\dfrac{dy}{dx}=\dfrac{1}{2-\frac{y}{x}}\\\sf fac_{\!\!,}a\\\sf v=\dfrac{y}{x}\implies y=v\cdot x\\\sf\dfrac{dy}{dx}=\dfrac{d}{dx}(v\cdot x)=x\dfrac{dv}{dx}+v\\\sf x\dfrac{dv}{dx}+v=\dfrac{1}{2-v}\\\sf x\dfrac{dv}{dx}=\dfrac{1}{2-v}-v\\\sf x\dfrac{dv}{dx}=\dfrac{1-2v+v^2}{2-v}\end{array}}$

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

$\boxed{\begin{array}{l}\sf\dfrac{2-v}{v^2-2v+1}=\dfrac{2-v}{(v-1)^2}=\dfrac{A}{v-1}+\dfrac{B}{(v-1)^2}\\\sf\dfrac{2-v}{(v-1)^2}=\dfrac{A(v-1)+B}{(v-1)^2}\\\sf 2-v=Av-A+B\\\sf 2-v=(B-A)+Av\\\begin{cases}\sf B-A=2\\\sf A=-1\end{cases}\\\sf B-(-1)=2\\\sf B+1=2\\\sf B=2-1\\\sf B=1\\\sf\dfrac{2-v}{(v-1)^2}=-\dfrac{1}{v-1}+\dfrac{1}{(v-1)^2}\end{array}}$

$\boxed{\begin{array}{l}\underline{\rm substituindo~na~equac_{\!\!,}\tilde ao~diferencial~temos:}\\\sf\bigg(-\dfrac{1}{v-1}+\dfrac{1}{(v-1)^2}\bigg)dv=\dfrac{dx}{x}\\\underline{\rm integrando~ambos~lados~temos:}\\\displaystyle\sf\int\bigg(\dfrac{-1}{v-1}+\dfrac{1}{(v-1)^2}\bigg)dv=\int\dfrac{dx}{x}\\\displaystyle\sf-\int\dfrac{dv}{v-1}+\int\dfrac{dv}{(v-1)^2}=\int\dfrac{dx}{x}\\\sf-\ell n|v-1|-\dfrac{1}{v-1}=\ell n|x|+c\\\sf -\ell n\bigg|\dfrac{y}{x}-1\bigg|-\dfrac{1}{\frac{y}{x}-1}=\ell n|x|+c\end{array}}$

$\boxed{\begin{array}{l}\sf-\ell n\bigg|\dfrac{y-x}{x}\bigg|+\dfrac{x}{y-x}=\ell n|x|+c\end{array}}$

$\boxed{\begin{array}{l}\rm 2)~\sf xy'=y+\sqrt{x^2+4y^2},x>0\\\sf x\dfrac{dy}{dx}=y+\sqrt{x^2+4y^2}\\\sf\dfrac{dy}{dx}=\dfrac{y}{x}+\dfrac{\sqrt{x^2+4y^2}}{x}\\\sf\dfrac{dy}{dx}=\dfrac{y}{x}+\dfrac{\sqrt{x^2\bigg(1+4\cdot\bigg(\dfrac{y}{x}\bigg)^2\bigg)}}{x}\\\\\sf\dfrac{dy}{dx}=\dfrac{y}{x}+\dfrac{\backslash\!\!\!x\cdot\sqrt{1+4\cdot\bigg(\dfrac{y}{x}\bigg)^2}}{\backslash\!\!\!x}\\\sf\dfrac{dy}{dx}=\dfrac{y}{x}+\sqrt{1+4\cdot\bigg(\dfrac{y}{x}\bigg)^2}\end{array}}$

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

$\boxed{\begin{array}{l}\sf x\dfrac{dv}{dx}+\backslash\!\!\!v=\backslash\!\!\!v+\sqrt{1+4v^2}\\\sf\dfrac{dv}{dx}=\sqrt{1+4v^2}\\\sf\dfrac{dv}{\sqrt{1+4v^2}}=dx\\\underline{\rm integrando~em~ambos~lados~temos:}\\\displaystyle\sf\int\dfrac{dv}{\sqrt{1+4v^2}}=\int dx\end{array}}$

$\boxed{\begin{array}{l}\displaystyle\sf\int\dfrac{dv}{\sqrt{1+4v^2}}=\int\dfrac{dv}{\sqrt{4(\frac{1}{4}+v^2)}}=\dfrac{1}{2}\int\dfrac{dv}{\sqrt{\frac{1}{4}+v^2}}\\\sf =\dfrac{1}{2}\ell n\bigg|v+\sqrt{v^2+\dfrac{1}{4}}\bigg|+c\end{array}}$

$\boxed{\begin{array}{l}\underline{\rm substituindo~temos}\\\sf\dfrac{1}{2}\ell n\bigg|v+\sqrt{v^2+\dfrac{1}{4}}\bigg|=x+c\\\sf \ell n\bigg|v+\sqrt{v^2+\dfrac{1}{4}}\bigg|=2x+k\\\sf v+\sqrt{v^2+\dfrac{1}{4}}=ke^{2x}\\\sf\dfrac{y}{x}+\sqrt{\bigg(\dfrac{y}{x}\bigg)^2+\dfrac{1}{4}}=ke^{2x}\end{array}}$