Question

Seja y(x) a solução do problema de valor inicial y' + xy2 = x , y(0) = 0. Quanto vale y(1)?

Answer 1

$\frac{dy}{dx}+xy^2 = x\\\\\frac{dy}{dx}+xy^2-x = 0\\\\\frac{dy}{dx}+x(y^2-1)=0\\\\\frac{dy}{dx}=-x(y^2-1)\\\\\frac{1}{y^2-1}dy=-xdx\\\\\int\frac{1}{(y+1)(y-1)}dy=-\int xdx\\\\\int\frac{A(y-1)+B(y+1)}{(y+1)(y-1)}dy=-\frac{x^2}{2}+c\\\\A+B=0\\B-A=1\\\\A=-\frac{1}{2}\\\\B=\frac{1}{2}\\\\-\frac{1}{2}\int\frac{1}{y+1}dy+\frac{1}{2}\int\frac{1}{y-1}dy=-\frac{x^2}{2}+c\\\\ln|y-1|-ln|y+1|=-x^2+c\\\\ln|\frac{y-1}{y+1}|=-x^2+c\\\\\\y(0)=0\\\\ln|-1|=c\\\\c=0\\\\ln|\frac{y-1}{y+1}|=-x^2$

$|1-\frac{2}{y+1}|=e^{-x^2}\\\\-\frac{2}{y+1}=\pm e^{-x^2}-1\\\\\frac{1}{y+1}=\frac{1\pm e^{-x^2}}{2}\\\\y=\frac{2}{1\pm e^{-x^2}}-1\\\\y(1)=\frac{2}{1\pm \frac{1}{e}}-1\\\\\boxed{y(1)=1-\frac{2e}{1\pm e}}$

Analisando as alternativas temos:

$y_1(1)=1-\frac{2e}{1+e}\\\\y_1(1)=\frac{1+e-2e}{1+e}\\\\y_1(1)=\frac{1-e}{1+e}\\\\y_1(1) = \frac{e-1}{e+1}\\\\\\y_2(1)=1-\frac{2e}{1-e}\\\\y_2(1)=\frac{1-e-2e}{1-e}\\\\y_2(1)=\frac{1-3e}{1-e}\\\\y_2(1)=\frac{3e-1}{e-1}$

Então a alternativa correta, que coincide com o valor da primeira função válida do PVI é:

$\boxed{d)\ y(1)=\frac{e-1}{e+1}}$