Question

1. (a) Find the solution of the initial-value problem and determine its interval of existence (i.e domain of the resulting function y).
(i)dy =1+t2, y(t=1)=0. dt t
(ii)(t+1)dy =1−y, y(t=0)=3. [Hint: LetA=−(±e−C).] dt

Answer 1

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

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

(ii)

$(t+1)\frac{dy}{dt}=1-y\to\int(\frac{dy}{1-y})=\int(\frac{dt}{t+1})\to -\log(y-1)=\log(t+1)+C_1\to \\ y(t)=\frac{C_2}{t+1}+1; \, y(0)=3\to C_2 +1=3\to C_2 =2. \\Answer:y(t)=\frac{t+3}{t+1}. (i) y\in R; (ii) y\in(-\infty,1)\bigcup(1,+\infty)$