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

Rewrite as
\[\frac{dx}{2x+5}=\frac{dt}{t^2+2t}\]
\[\frac{dx}{2x+5}=\frac{dt}{t^2+2t+1-1}\]
\[\frac{dx}{2x+5}=\frac{dt}{(t+1)^2-1}\]
Integrate both sides

Answer 2

ok lemme see.

Answer 3

ok so now.. ln(2x + 5) = ln(t) - ln(t + 2)

Answer 4

that sound right?