austinL
  • austinL
Differential Equation, Initial Value Problem I have started to get how to do these fairly easily, however this one has me a tad confused.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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austinL
  • austinL
\(3y\prime -2y=e^{\frac{-\pi~t}{2}}\) \(y(0)=a\)
austinL
  • austinL
I have this so far, \(\Large{y\prime -\frac{2}{3}y=\frac{e^{\frac{\pi t}{2}}}{3}}\)
Loser66
  • Loser66
where is - sign

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austinL
  • austinL
I forgot it, excuse me.
Loser66
  • Loser66
apply formula, that's it
austinL
  • austinL
\(\Large{y\prime -\frac{2}{3}y=\frac{e^{\frac{-\pi t}{2}}}{3}}\) I guess it was just the right side that was confusing me.
Loser66
  • Loser66
rhs is g(t) , just apply formula to get the answer.
austinL
  • austinL
\(\Large{y=e^{\frac{2}{3}t}\int e^{-\frac{2}{3}t}~\frac{e^{\frac{\pi t}{2}}}{3}}dt+Ce^{\frac{2}{3}t}\) This look correct?
Loser66
  • Loser66
yep
Loser66
  • Loser66
ok, you are on the right track, just take integral, then replace y (0) to find C , then, plug back
austinL
  • austinL
Gimme a minute with this :P
austinL
  • austinL
Ok, I have a gnarly answer for the integral.
austinL
  • austinL
\(\Large{\frac{2e^{\frac{1}{6}(3\pi -4)t}}{3\pi -4}}\)
austinL
  • austinL
Whole answer, \(\Large{y=e^{\frac{2}{3}t}\times\frac{2e^{\frac{1}{6}(3\pi -4)t}}{3\pi -4}+Ce^{\frac{2}{3}t}}\)
Loser66
  • Loser66
look good to me, simplify a little bit
Loser66
  • Loser66
find c , too
Loser66
  • Loser66
@austinL knock knock, finish it, friend. You don't have the final answer yet.
austinL
  • austinL
I got this, \(\Large{y(t)=\dfrac{2e^{\frac{1}{6}(3\pi -4)t+\frac{2t}3}}{3\pi-4} +Ce^{\frac{2}{3}t}}\)
austinL
  • austinL
\(\Large{y(t)=\dfrac{2e^{\frac{1}{6}(3\pi -4)t+\frac{2t}{3}}}{3\pi -4}+(a-\dfrac{2}{3\pi-4})e^{\frac{2}{3}t}}\) Final answer.
austinL
  • austinL
Look ok @amistre64 ?
amistre64
  • amistre64
\[\Large{y(t)=\dfrac{2e^{\frac{1}{6}(3\pi -4)t+\frac{2t}3}}{3\pi-4} +Ce^{\frac{2}{3}t}}\] \[\Large{y(0)=a=\dfrac{2e^{\frac{1}{6}(0+0)}}{3\pi-4} +Ce^{0}}\] \[\Large{a=\dfrac{2}{3\pi-4} +C}\] as long as the step up to it are fine :) looks ok to me
austinL
  • austinL
Very cool, thanks!

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