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

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{} \frac{t+1}{\sqrt{t}} dt\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0split the fraction in two and work each side

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0{S} t.(t^1/2) + t^(1/2) dt {S} t^(1/2) + t^(1/2) dt

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0t^(3/2) t^(1/2)  +  + C 3/2 1/2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0when t = 1; this should equal 4 according to the initial condition

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.02/3 + 2 +C = 4 8/3 +C = 4 C = 4  8/3 = 16/3 if i did it right

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{2t^2}{3}+\frac{2\sqrt{t}}{1}\frac{16}{3}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ahh 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
 5 years ago
Best ResponseYou've already chosen the best response.0though solving for c = 4/3, but otherwise it's golden

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, the integrating i can do.... addition? nah lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0haha it's cool. i'm usually the same way just apparently brain dead tonight
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