Question

21, 22, 23, 24, 25, 26, 27, and 28 solve the initial - value problem.

1. $\frac{dy}{dt}=t^{2}+1$, $tgeq0$, $y = 6$ when $t = 0$
2. $\frac{dy}{dt}=1+\frac{2}{t}$, $t>0$, $y = 5$ when $t = 1$

Explanation:

Step1: Integrate the right - hand side

Integrate $\frac{dy}{dt}=t^{2}+1$ with respect to $t$. Using the power rule $\int t^{n}dt=\frac{t^{n + 1}}{n+1}+C$ ($n
eq - 1$), we have $y=\int(t^{2}+1)dt=\frac{t^{3}}{3}+t + C$.

Step2: Find the constant $C$

Substitute $t = 0$ and $y = 6$ into $y=\frac{t^{3}}{3}+t + C$. We get $6=\frac{0^{3}}{3}+0 + C$, so $C = 6$.

for 22:

Step1: Integrate the right - hand side

Integrate $\frac{dy}{dt}=1+\frac{2}{t}$ with respect to $t$. We know that $\int 1dt=t$ and $\int\frac{2}{t}dt=2\ln|t|$ (since $t>0$, we can write it as $2\ln t$). So $y=t + 2\ln t+C$.

Step2: Find the constant $C$

Substitute $t = 1$ and $y = 5$ into $y=t + 2\ln t+C$. We have $5=1+2\ln(1)+C$. Since $\ln(1)=0$, then $5=1 + 0+C$, so $C = 4$.

Answer:

$y=\frac{t^{3}}{3}+t + 6$