Answer to Question #124100 in Calculus for Koby

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) $\frac{dy}{dt}=\frac{1+t^2}{t} , y(t = 1) = 0.$

We can separate variables and integrate this equation:

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

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

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

$y(t)=ln|t|+\frac{t^2}{2}+C,$ $C$ is a constant.

Next, we plug in our initial condition: $y(t=1)=0=ln|1|+\frac{1^2}{2}+C=0.5+C.$

We get that $C=-0.5.$

Thus, plugging back into our solution for y, we get:

$y(t)=ln|t|+\frac{t^2}{2}-0.5.$

It's domain is $t\neq 0$ or $t \in (-\infty;0)\cup(0;+\infty)$ .

(ii) $(t+1)\frac{dy}{dt} = 1- y , y(t = 0) = 3.$

We can separate and integrate:

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

$-\frac{d(1-y)}{1-y}=\frac{d(1+t)}{dt},$

$−ln∣1−y∣=ln∣1+t∣+C,$ $C$ is a constant.

$ln|(1-y)(1+t)|=-C,$

$(1-y)(1+t)=e^{-C}.$

Set $A=e^{-C},$ hence we have:

$(1−y)(1+t)=A,$ where A is a constant.

$1-y=\frac{A}{1+t},$

$y=1-\frac{A}{1+t}=\frac{1+t-A}{1+t}=\frac{t+C_1}{t+1},$ where $C_1=1-A$ is a constant.

Next, we plug in our initial condition:

$y(t=0)=3=\frac{0+C_1}{0+1}=C_1,$ and we get that $C_1=3.$

Thus, plugging back into our solution for y, we get:

$y(t)=\frac{t+3}{t+1},$ it's domain is $t \neq -1$ or $t \in (-\infty; -1)\cup (-1; +\infty).$