Question

@amistre64 Solve the initial value problem below:

(t^2 + 2t)dx/dt = 2x + 5 with t, x>0 and x(1) = 1

Answer 1

Why do you keep reposting it
it was solved for x
\[x=\frac{7\sqrt3}{2}(\frac{t}{t+2})-5\]

Answer 2

that isn't correct.. :/

Answer 3

Sorry correction
\[x=\frac{7}{3√2}\sqrt{(\frac{t}{t+2})}−5\]

Answer 4

ahhh, let me try that.

Answer 5

that's not right either..

Answer 6

system froze ...

Answer 7

uh oh.. not good.

Answer 8

make sure your question is posted correctly; no typos please

Answer 9

ok I'll screen shot it.

Answer 10

IT is
\[\frac{7\sqrt3}{2}\]
it was an typing error

Answer 11

it still says that is incorrect

Answer 12

aw it also -5/2
I am sorry i usually mess up the numbers

Answer 13

No worriest at all. let me try that.

Answer 14

by 5/2 do you mean.. ( 7/(3sqrt(2)) * sqrt( t / (t+2)) -5/2)?

Answer 15

Y

Answer 16

didn't work either... :/

Answer 17

\[(2x + 5)^{-1}dx = (t^2 + 2t)^{-1} dt \]
\[\frac 12 ln(2x + 5) = \int (t^2 + 2t)^{-1} dt \]
gotta partial that up i think

1 A B
----- = -- + ---
t(t+2) t t+2

1 = A(t+2) + Bt ; when t=0; A=1/2
when t=-2 B=-1/2

\[\frac 12 ln(2x + 5) = \int \frac{1/2}{t}-\frac{1/2}{t+2} dt \]
\[\frac 12 ln(2x + 5) = \frac{1}{2}ln(t)-\frac{1}{2}ln(t+2) +C \]
\[ln(2x + 5) = ln(t)-ln(t+2) +C \]
\[ln(2x + 5) = ln(\frac{t}{t+2}) +C \]
\[2x + 5 = \frac{Ct}{t+2} \]
maybe

Answer 18

Aw i messed up the lhs @amistre64 is right

Answer 19

so if its good; x(1) = 1

\[7 = \frac{C}{3}\]
C = 21

Answer 20

\[x = \frac{21t}{2(t+2)}-\frac52\]
but id have to check the wolf

Answer 21

That sir, is correct. you rock. NotSObright, thanks for your help alot too bud.

Answer 22

yay!!