Question

ALGUEM ME AJUDA NESSA QUESTÃO PF???? é pra hj

!Resolver a equação diferencial homogênea usando o método da substituição!

Answer 1

$\boxed{\begin{array}{l}\sf\dfrac{dy}{dx}=\dfrac{x-y}{x}\\\sf\dfrac{dy}{dx}=\dfrac{\backslash\!\!\!x\cdot\bigg(1-\dfrac{y}{x}\bigg)}{\backslash\!\!\!x}\\\sf\dfrac{dy}{dx}=1-\dfrac{y}{x}\\\underline{\rm fac_{\!\!,}a}\\\sf v=\dfrac{y}{x}\implies y=v\cdot x\\\sf \dfrac{dy}{dx}=\dfrac{d}{dx}(v\cdot x)\\\sf \dfrac{dy}{dx}=x\dfrac{dv}{dx}+v\\\underline{\rm substituindo~na~equac_{\!\!,}\tilde ao~diferencial~temos:}\\\sf x\dfrac{dv}{dx}+v=1-v\\\sf v+v-1=-x\dfrac{dv}{dx}\\\sf 2v-1=-x\dfrac{dv}{dx}\end{array}}$

$\boxed{\begin{array}{l}\sf\dfrac{dv}{2v-1}=-\dfrac{dx}{x}\\\displaystyle\sf\int\dfrac{dv}{2v-1}=-\int\dfrac{dx}{x}\\\underline{\rm c\acute alculo~da~integral~da~esquerda:}\\\underline{\rm fac_{\!\!,}a}\\\sf u=2v-1\implies dv=\dfrac{1}{2}du\\\displaystyle\sf\int\dfrac{dv}{2v-1}=\dfrac{1}{2}\int\dfrac{du}{u}=\dfrac{1}{2}\ell n|u|+k\\\displaystyle\sf\int\dfrac{dv}{2v-1}=\dfrac{1}{2}\ell n|2v-1|+k_1\end{array}}$

$\Large\boxed{\begin{array}{l}\underline{\rm voltando~a~express\tilde ao~temos:}\\\displaystyle\sf\int\dfrac{dv}{2v-1}=-\int\dfrac{dx}{x}\\\sf\dfrac{1}{2}\ell n|2v-1|+k_1=-\ell n|x|+k_2\\\sf \ell n|(2v-1)^{\frac{1}{2}}|=-\ell n|x|+k\\\underline{\rm aplicando~\it ~e~\rm dos~dois~lados~temos:}\\\sf e^{\ell n|(2v-1)^{\frac{1}{2}}|}=e^{\ell n|\frac{1}{x}|+k}\\\sf (2v-1)^{\frac{1}{2}}=k\cdot\dfrac{1}{x}\\\underline{\rm substituindo~v~por~\dfrac{y}{x}~temos:}\end{array}}$

$\Large\boxed{\begin{array}{l}\sf\bigg(2\dfrac{y}{x}-1\bigg)^{\frac{1}{2}}=k\cdot\dfrac{1}{x}\\\sf\bigg(\dfrac{2y-x}{x}\bigg)^{\frac{1}{2}}=k\cdot\dfrac{1}{x}\\\sf\bigg[\bigg(\dfrac{2y-x}{x}\bigg)^{\frac{1}{2}}\bigg]^2=k\cdot\bigg(\dfrac{1}{x}\bigg)^2\\\sf\dfrac{2y-x}{\backslash\!\!\!x}=k\cdot\dfrac{1}{\backslash\!\!\!\!x^2}\\\sf 2y-x=k\cdot\dfrac{1}{x}\\\sf 2y=x+k\cdot\dfrac{1}{x}\\\\\sf y(x)=\dfrac{1}{2}\cdot\bigg[\dfrac{x^2+k}{x}\bigg]\end{array}}$