Question

Ache a solução diferencial y¹=-2xy², que atende a condição inicial y(0)=5e assinale a alternativa que contenha:

Answer 1

$\displaystyle \frac{dy}{dx} = -2xy^{2} \\ \\ \\ \frac{dy}{y^{2}} = -2x \, dx \\ \\ \\ \frac{1}{y^{2}} \, dy = -2x \, dx \\ \\ \\ \int \frac{1}{y^{2}} \, dy = \int -2x \, dx \\ \\ \\ - \frac{1}{y} + c' = -x^{2} + c'' \\ \\ \\ -\frac{1}{y} = -x^{2} + c'' - c' \\ \\ \\ -\frac{1}{y} = -x^{2} + c \\ \\ \\ \frac{1}{y} = x^{2} - c \\ \\ \\ y = \frac{1}{x^{2}-c}$

Agora vamos descobrir o valor da nossa constante.

$\displaystyle \frac{1}{0^{2}-c}=5 \\ \\ \\ \frac{1}{-c}=5 \\ \\ \\ -c=\frac{1}{5} \\ \\ \\ c=-\frac{1}{5}$

Acompanhe:

$\displaystyle y = \frac{1}{x^{2}-c} \\ \\ \\ y=\frac{1}{x^{2}-\displaystyle(- \frac{1}{5})} \\ \\ \\ y=\frac{1}{x^{2}+\displaystyle \frac{1}{5}} \\ \\ \\ y=\frac{1}{\displaystyle \frac{5x^{2}+1}{5}} \\ \\ \\ \boxed{\boxed{y=\frac{5}{x^{2}+1}}}$

Answer 2

y'=-2xy²

dy/dx = -2xy²

(1/y²) dy  = -2x dx


∫ (1/y²) dy  = -2 ∫ x dx

y⁻¹/(-1) =-2 * x²/2 + c  

y⁻¹ = x² + c    ...  y(0)=5

5⁻¹ = 0² + c  ==>c=1/5

y⁻¹ = x² + 1/5

1/y = (5x²+1)/5

y = 5/(5x²+1)