Question

Differential Equation, Initial Value Problem
I have started to get how to do these fairly easily, however this one has me a tad confused.

Answer 1

\(3y\prime -2y=e^{\frac{-\pi~t}{2}}\)
\(y(0)=a\)

Answer 2

I have this so far,

\(\Large{y\prime -\frac{2}{3}y=\frac{e^{\frac{\pi t}{2}}}{3}}\)

Answer 3

where is - sign

Answer 4

I forgot it, excuse me.

Answer 5

apply formula, that's it

Answer 6

\(\Large{y\prime -\frac{2}{3}y=\frac{e^{\frac{-\pi t}{2}}}{3}}\)

I guess it was just the right side that was confusing me.

Answer 7

rhs is g(t) , just apply formula to get the answer.

Answer 8

\(\Large{y=e^{\frac{2}{3}t}\int e^{-\frac{2}{3}t}~\frac{e^{\frac{\pi t}{2}}}{3}}dt+Ce^{\frac{2}{3}t}\)
This look correct?

Answer 9

yep

Answer 10

ok, you are on the right track, just take integral, then replace y (0) to find C , then, plug back

Answer 11

Gimme a minute with this :P

Answer 12

Ok, I have a gnarly answer for the integral.

Answer 13

\(\Large{\frac{2e^{\frac{1}{6}(3\pi -4)t}}{3\pi -4}}\)

Answer 14

Whole answer,
\(\Large{y=e^{\frac{2}{3}t}\times\frac{2e^{\frac{1}{6}(3\pi -4)t}}{3\pi -4}+Ce^{\frac{2}{3}t}}\)

Answer 15

look good to me, simplify a little bit

Answer 16

find c , too

Answer 17

@austinL knock knock, finish it, friend. You don't have the final answer yet.

Answer 18

I got this,

\(\Large{y(t)=\dfrac{2e^{\frac{1}{6}(3\pi -4)t+\frac{2t}3}}{3\pi-4} +Ce^{\frac{2}{3}t}}\)

Answer 19

\(\Large{y(t)=\dfrac{2e^{\frac{1}{6}(3\pi -4)t+\frac{2t}{3}}}{3\pi -4}+(a-\dfrac{2}{3\pi-4})e^{\frac{2}{3}t}}\)

Final answer.

Answer 20

Look ok @amistre64 ?

Answer 21

\[\Large{y(t)=\dfrac{2e^{\frac{1}{6}(3\pi -4)t+\frac{2t}3}}{3\pi-4} +Ce^{\frac{2}{3}t}}\]
\[\Large{y(0)=a=\dfrac{2e^{\frac{1}{6}(0+0)}}{3\pi-4} +Ce^{0}}\]
\[\Large{a=\dfrac{2}{3\pi-4} +C}\]
as long as the step up to it are fine :) looks ok to me

Answer 22

Very cool, thanks!