Question

**Calc 2 Help Needed** Solve the initial value problem below:

(t^2 + 2t)dx/dt = 2x + 5 with t, x>0 and x(1) = 1

Answer 1

rearrange terms to separate the x variables from t variables
\[\frac{dx}{2x+5} = \frac{dt}{t^{2}+2t}\]
integrate both sides
\[\frac{1}{2}\ln (2x+5) = \frac{1}{2}(\ln t -\ln (t+2) +C)\]
Let C = ln(k) , combine logs
\[\ln(2x+5) = \ln(\frac{kt}{t+2})\]
\[2x+5 = \frac{kt}{t+2}\]
\[x = \frac{kt}{2(t+2)}-\frac{5}{2}\]
solve for k using x(1)=1
\[1 = \frac{k}{6}-\frac{5}{2}\]
\[k = 21\]

\[x(t) = \frac{21t}{2(t+2)}-\frac{5}{2}\]