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

(t^2 + 2t)dx/dt = 2x + 5 with t, x0 and x(1) = 1

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

dumbcow · Answer

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
$$1 = \frac{k}{6}-\frac{5}{2}$$
$$k = 21$$

$$x(t) = \frac{21t}{2(t+2)}-\frac{5}{2}$$