Question

Solução geral da equaçao diferencial dy/dx = √xy

Answer 1

EDO separável:
$\boxed{\frac{dy}{dx}=f(x)\implies y=\int f(x)dx}$

$\displaystyle i)~~~~\frac{dy}{dx}=\sqrt{xy}\\\\ii)~~~\frac{dy}{\sqrt{y}}=\sqrt{x}dx\\\\iii)~~\int\frac{dy}{\sqrt{y}}=\int\sqrt{x}dx\\\\~~~~~\sqrt{y}=u\implies du=\frac{1}{2u}dy\implies dy=2udu\\\\~~~~~\int\frac{dy}{\sqrt{y}}=2\int\frac{u}{u}du=2\int du=2u+c_1=2\sqrt{y}+c_1\\\\~~~~\int \sqrt{x}dx=\frac{2}{3}\sqrt{x}^3+c_2\\\\iv)~~~2\sqrt{y}=\frac{2}{3}\sqrt{x}^3+A\\\\v)~~\sqrt{y}=\frac{\sqrt{x}^3}{3}+\frac{A}{2}\\\\vi)~~y=\left(\frac{\sqrt{x}^3}{3}+\frac{A}{2}\right)^2\\\\vii)~~\boxed{\boxed{y(x)=\frac{x^3}{9}+\frac{\sqrt{x}^3A}{3}+\frac{A^2}{4}}}}$

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Answer 2

$y'=\sqrt{xy}\ \to\ \frac{y'}{\sqrt{y}}=\sqrt{x}\ \\ \frac{1}{\sqrt{y}}.\frac{dy}{dx}=\sqrt{x}\ \to\ \frac{1}{\sqrt{y}}.dy=\sqrt{x}.dx\\\\ Integre\ em\ ambos\ os\ lados:\\\\ \int \frac{1}{\sqrt{y}}.dy=\int \sqrt{x}.dx\ \to\ \int y^{\frac{-1}{2}}.dy=\int x^{\frac{1}{2}}.dx\\\\ 2\sqrt{y}+c_1=\frac{2x^{\frac{3}{2}}}{3}+c_2\ \to\ 2\sqrt{y}= \frac{2x^{\frac{3}{2}}}{3}+c$

$Multiplique\ por\ 3\ em\ ambos\ os\ lados: \\\\ 3.(2\sqrt{y})=3.(\frac{2x^{\frac{3}{2}}}{3}+c)\ \to\ 6\sqrt{y}=2x^{\frac{3}{2}}+3c\\\\ Eleve\ ao\ quadrado\ em\ ambos\ os\ lados:\\\\ (6\sqrt{y})^2=(2x^{\frac{3}{2}}+3c)^2\\\\ 36y=4x^3+12x^{\frac{3}{2}}c+9c^2\\\\ y= \frac{4x^3+12x^{\frac{3}{2}}c+9c^2}{36}\\\\ \| \ y=\frac{x^3}{9}+\frac{(x\sqrt{x}).c}{3}+4c^2\ \|$