Question

5) Resolver a equação diferencial homogênea 2y - xy' =0
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Answer 1

Explicação passo-a-passo:

$\\ \displaystyle \mathsf{2y - xy' = 0} \\\\ \mathsf{x \ \frac{dy}{dx} = 2y} \\\\ \mathsf{\frac{1}{y} dy = \frac{2}{x} dx} \\\\ \mathsf{\int \frac{1}{y} dy = \int \frac{2}{x} dx} \\\\ \mathsf{\ln |y| = 2 \cdot \ln |x| + c} \\\\ \mathsf{\ln |y| - \ln x^2 = c} \\\\ \mathsf{\ln \frac{|y|}{x^2} = c} \\\\ \mathsf{e^c = \frac{y}{x^2}} \\\\ \mathsf{y = e^c \cdot x^2} \\\\ \boxed{\boxed{\mathsf{y = Cx^2}}}$

Answer 2

$2y - xy' =0\\xy'=2y$

$x\frac{dy}{dx}=2y \\\frac{dy}{y}=2\frac{dx}{x}$

$\int\frac{dy}{y}=2\int\frac{dx}{x}$

$ln|y|=2ln|x|+c$

${e}^{ln(y)}={e}^{2ln(x) +c}$

${e}^{ln(y)} ={e}^{2ln(x)}.{e}^{c}$

$\boxed{\boxed{y=k{x}^{2}}}$