## anonymous 4 years ago does the IVP dy/dt=y*t^(1/2), y(1)=1 have a unique solution on an interval containing t=1?

1. amistre64

looks seperable to me

2. amistre64

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

3. anonymous

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

4. amistre64

$\large y =exp(\frac{2}{3}t^{3/2}+C)$ $y =exp(\frac{2}{3}t^{3/2})*exp(C)$ $y =C_1\ exp(\frac{2}{3}t^{3/2})$ $1 =C_1\ exp(\frac{2}{3})$ $exp(-\frac23) =C_1$ if i see it right

5. anonymous

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

6. amistre64

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

7. mathmate

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

8. amistre64

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

9. anonymous

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

10. amistre64

divide off the e^{2/3}

11. amistre64

division makes an exponent go negative

12. anonymous

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

13. anonymous

ok

14. amistre64

yes

15. anonymous

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...