Question

does the IVP dy/dt=y*t^(1/2), y(1)=1 have a unique solution on an interval containing t=1?

Answer 1

looks seperable to me

Answer 2

\[ln|y| =\frac{2}{3}t^{3/2}+C \]

Answer 3

IVP=initial value problem. i have y=e^(2/3t^(3/2))+1-e^(2/3) but it should be y=e^(2/3(t^(3/2)-1))...not the same

Answer 4

\[\large y =exp(\frac{2}{3}t^{3/2}+C)\]
\[y =exp(\frac{2}{3}t^{3/2})*exp(C)\]

\[y =C_1\ exp(\frac{2}{3}t^{3/2})\]
\[1 =C_1\ exp(\frac{2}{3})\]
\[exp(-\frac23) =C_1\]
if i see it right

Answer 5

you say seperable meaning not unique? cuz it should be unique...

Answer 6

seperable meaning its easy to see if theres a solution :)

Answer 7

The solution is unique within an interval if the function y and its derivative are continuous on the interval.

Answer 8

\[\frac{dy}{dx}=f(x)\ h(y)\]
\[\frac{1}{h(y)}dy=f(x)dx\]
\[\int(\frac{1}{h(y)}dy=f(x)dx)\]

Answer 9

i see where i messed up. but why do you have -2/3?^^^

Answer 10

divide off the e^{2/3}

Answer 11

division makes an exponent go negative

Answer 12

is that 1/e^(2/3)?

Answer 13

ok

Answer 14

yes

Answer 15

same question...\[dy/dt=6y ^{2/3}\] y(1)=0. i'm not getting 0 so i need verification. i got \[y=(2t-C)^{3}\] C=-8/19...