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 , (ii) (t + 1)dy y(t = 1) = 0. dt = 1−y, y(t = 0) = 3. [Hint: Let A = −(±e−C).]

Answer 1

i).Given,

$$\frac{dy}{dt}=1+t^2$$

Initial condition is not given,thus we are solving the general solution.

$$dy=(1+t^2)dt\\ \implies \int dy=\int(1+t^2)dt\\\implies y=t+\frac{t^3}{3}+C$$

Where, $C$ is constant.

Clearly, $y$ is a family of cubic polynomial in $t$ ,thus domain is $\mathbb{R}$

ii).Question is badly stated.