solve the initial value problem du/dt = (t+1)/sqrt(t); u(1) = 4

Question

amistre64 · Answer

$$\int\limits_{} \frac{t+1}{\sqrt{t}} dt$$

amistre64 · Answer

split the fraction in two and work each side

amistre64 · Answer

{S} t.(t^-1/2) + t^(-1/2) dt

{S} t^(1/2) + t^(-1/2) dt

amistre64 · Answer

t^(3/2)   t^(1/2)
----- + ------ + C
  3/2         1/2

amistre64 · Answer

when t = 1; this should equal 4 according to the initial condition

amistre64 · Answer

2/3 + 2 +C = 4

8/3 +C = 4

C = 4 - 8/3 = -16/3  if i did it right

amistre64 · Answer

$$\frac{2t^2}{3}+\frac{2\sqrt{t}}{1}-\frac{16}{3}$$

anonymous · Answer

ahh thanks a ton.  i was WAY overthinking it.  used to other profs who constantly used all kinds of tricks and stuff so i was looking for something way more complicated.  thanks again.

anonymous · Answer

though solving for c = 4/3, but otherwise it's golden

amistre64 · Answer

:)  youre welcome

amistre64 · Answer

yeah, the integrating i can do.... addition?  nah lol

anonymous · Answer

haha it's cool.  i'm usually the same way- just apparently brain dead tonight