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

(t^2 + 2t)dx/dt = 2x + 5, multiply both sides by dt:
(t^2 + 2t)dx = (2x + 5)dt, divide both sides by t^2+2t
dx = (2x + 5)dt/(t^2 + 2t), divide both sides by 2x+5
dx/(2x+5) = dt/(t^2+2t)

Answer 2

Integrate both sides:
Integral(dx/(2x+5)) = Integral(dt/(t^2+2t))
For the left integral use substitution u = 2x+5, du = 2xdx
For the right integral use partial fractions: A/t + B/(t+2) = 1/(t*(t+2))

Answer 3

solve for both integrals and tell me what u get. If u get lost, ask. I'll be working on my own hw meanwhile...

Answer 4

YOu still there??