Question

Encontrando a solução do problema de valor iniciaL y'+2y=te−2t, y(1)=0 obtemos:


a) y = (t−1)e−2t2

b) y = (t2−1)e−2t2

c) y = (t2−1)et

d) y = (t2−1)e2t

e) y = (t2−1)e−2t

Answer 1

Resposta: B) y = (t2−1)e−2t2