Question

Find the solution of the initial-value problem and determine its interval of existence (i.e domain of the resulting function y).

(i) dy

dt = 1+t2

t ; y(t = 1) = 0:

Answer 1

$\frac{dy}{dt}=1+t^2\to\ y(t)=\int(t^2+1)dt\to y(t)=\frac{1}{3}t^3+t+C_1.$

We take into account the initial conditions $y(1)=0\to C_1 +\frac{4}{3}=0\to C_1=-\frac{4}{3}.$

$Substitute C_1 =-\frac{4}{3}$ into $y(t)=\frac{t^3}{3}+t+C_1. Answer: y(t)=\frac{1}{3}(t^3+3t-4).$ y∈R