Question

resolva a equação diferencial y'-2xy=2x

Answer 1

dy/dx-2xy=2x

dy/dx =2x*(1+y)

1/(1+y)*  dy =2x * dx

∫ 1/(1+y) *  dy = ∫ 2x * dx

*** ∫ 1/(1+y) *  dy  ..fazendo 1+y=u  ==>dy=du

***∫ 1/u * du = ln u

***como u= 1+y  ==>  ∫ 1/(1+y) *  dy  = ln (1+y)

∫ 1/(1+y) *  dy = ∫ 2x * dx

ln (1+y) =2x²/2 + c

1+y = e ^(x²+c)

1+y =e^(x²) * e^c ..........fazendo e^c=c₁

1+y =c₁ * e^(x²)

y= c₁*e^(x²) -1     ...c₁ é uma constante 

Answer 2

Resposta:

$\boxed{\mathtt{y = Ce^{x^2} - 1}}$

Explicação passo-a-passo:

$\\ \displaystyle \mathsf{y' - 2xy = 2x} \\\\ \mathsf{\frac{dy}{dx} = 2xy + 2x} \\\\\\ \mathsf{\frac{dy}{dx} = 2x \cdot (y + 1)} \\\\\\ \mathsf{\frac{1}{y + 1} \, dy = 2x \, dx} \\\\\\ \mathsf{\int \frac{1}{y + 1} \, dy = \int 2x \, dx}$

$\\ \displaystyle \mathsf{\ln \left | y + 1 \right | = x^2 + c} \\\\ \mathsf{e^{x^2 + c} = |y + 1|} \\\\ \mathsf{y + 1 = e^{x^2} \cdot e^c, \qquad \qquad - 1 < y < \infty} \\\\ \boxed{\boxed{\mathsf{y = Ce^{x^2} - 1}}}$