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

Answer 1

(i) Given

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

integrating both sides,

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

solving it, we obtain

$y = lnt + \frac{1}{2}t^2 +C$

applying the condition, y(t=1) = 0

$c = -\frac{1}{2}$

so equation will be,

$y = lnt + \frac{1}{2}t^2 -\frac{1}{2}$

Domain of y is $[0,\infin)$

(ii) Given $(t+1)\frac{dy}{dt} = 1-y$

integrating on both sides,

$\int\frac{dy}{1-y} = \int\frac{dt}{1+t}$

$-ln(1-t) = ln (1+t)+lnC$

applying the condition, y(t=0) = 3

$C = -\frac{1}{2}$

so equation will be,

$y = 1 + \frac{2}{1+t} = \frac{t+3}{t+1}$

Domain for y is $\R$ - {1} where R is real number.