anonymous
  • anonymous
does the IVP dy/dt=y*t^(1/2), y(1)=1 have a unique solution on an interval containing t=1?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
looks seperable to me
amistre64
  • amistre64
\[ln|y| =\frac{2}{3}t^{3/2}+C \]
anonymous
  • 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

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amistre64
  • 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
anonymous
  • anonymous
you say seperable meaning not unique? cuz it should be unique...
amistre64
  • amistre64
seperable meaning its easy to see if theres a solution :)
mathmate
  • mathmate
The solution is unique within an interval if the function y and its derivative are continuous on the interval.
amistre64
  • amistre64
\[\frac{dy}{dx}=f(x)\ h(y)\] \[\frac{1}{h(y)}dy=f(x)dx\] \[\int(\frac{1}{h(y)}dy=f(x)dx)\]
anonymous
  • anonymous
i see where i messed up. but why do you have -2/3?^^^
amistre64
  • amistre64
divide off the e^{2/3}
amistre64
  • amistre64
division makes an exponent go negative
anonymous
  • anonymous
is that 1/e^(2/3)?
anonymous
  • anonymous
ok
amistre64
  • amistre64
yes
anonymous
  • 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...

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