Question

Dada a equação x dy/dx -2y=6x, com valor inicial (2) = 0 entao y (3) e igual a

Answer 1

$x* \frac{dy}{dx} -2y = 6x\\\\ \frac{dy}{dx} - \frac{2y}{x}=6 \\\\\text{encontrando o fator integrante}\\\\ \mu = e^{ \int\frac{-2}{x}\;dx}\\\\ \mu = e^{-2ln(x)}\\\\\mu = x^{-2} \\\\\mu = \frac{1}{x^2} \\\\ \text{ multiplicando a equacao pelo fator integrante}\\\\ \frac{dy}{dx}*\frac{1}{x^2} - 2y *\frac{1}{x^3} = \frac{6}{x} \\\\\text{quando vc faz isso acaba encontrando uma derivada do produto}$

$\boxed{\boxed{\frac{dy}{dx}*\frac{1}{x^2} - 2y *\frac{1}{x^3} \to \frac{d}{dx}\left(y* \frac{1}{x^2} \right) }}\\\\\\ \text{entao temos }\\\\ \frac{d}{dx}\left(y* \frac{1}{x^2} \right) = \frac{6}{x^2} \\\\ \left(y* \frac{1}{x^2} \right )= \int \frac{6}{x^2 }\;dx \\\\ y* \frac{1}{x^2} = \frac{-6}{x} +C\\\\ \boxed{\boxed{y= -6x +Cx^2}}\\\\\ y(2)=0\\\\ 0= -6*2 +C*2^2\\\\ \frac{12}{4}=C \\\\ 3=C\\\\\\\boxed{\boxed{y(x)=-5x+3x^2}}$


calculando y(3)

$y(3)=-5*3+3*3^2 = 12$