Question

whats next

Answer 1

\[\huge \rm t \cdot \frac{dQ}{dt} +2Q =e^t ; Q(1)=2\]
\[\large\rm \frac{ dQ }{ dt } =(\frac{ e^t}{t}-\frac{2Q}{t}) \]
\[\large\rm dQ =(\frac{ e^t}{t}-\frac{2Q}{t}) dt\]\[\large\rm \int\limits_{}^{} dQ = \int(\frac{ e^t}{t}-\frac{2Q}{t}) dt\]

Answer 2

Hmm I don't think this one is separable :(
Are you familiar with `integrating factor` technique?

Answer 3

whats that

Answer 4

Oh noooo I just lost what I was writing T.T Misclick lol

Answer 5

X"D

Answer 6

\[\large\rm Q'+\frac{2}{t}Q=\frac{e^t}{t}\]
So ummm, remember your product rule,
I'll use Q just to show you what we're sort of looking for,\[\large\rm (uQ)'=uQ'+u'Q\]

So what we want to do is,
multiply the left side of our equation by some special value to make it look like an expanded product rule. From there, we can simplify it to look like left side of product rule equation.

Answer 7

Q_Q

Answer 8

You make this special substitution,
we call this our integrating factor,\[\large\rm u=e^{\int\limits P(t)dt}\]Where P(t) is the thing on the Q, so in this case, 2/t.\[\large\rm u=e^{\int\limits \frac2t~dt}\]

I can show you where this weird integrating factor comes from if you care,
it's not too difficult to understand I think :o

So anyway, for our problem,\[\large\rm u=e^{2\ln t}\qquad\to\qquad u=e^{\ln t^2}\qquad\to\qquad u=t^2\]

Answer 9

So if we multiply through by t^2,
it will turn out differential equation into an expanded product rule that we can manipulate.\[\large\rm t^2\left(Q'+\frac2tQ\right)=t^2\left(\frac{e^t}{t}\right)\]

Answer 10

\[\large\rm t^2Q'+2tQ=te^t\]

And on the left side, you can see that 2t is the derivative of t^2, right?
So our left side is actually,\[\large\rm (t^2Q)'\]

Answer 11

Too confusing? :U
What do you think?

Answer 12

i think milkyway > twix

Answer 13

What?? That's crazy talk!

Butterfinger >>> all

Answer 14

sorry

Answer 15

\[\large\rm t^2Q'+2tQ=te^t\]\[\large\rm (t^2Q)'=te^t\]Integrate both sides,\[\large\rm t^2Q=\int\limits te^t~dt\]and then integration by parts, ya?

Oh well XD if it's too much I sorry

Answer 16

ok so there is 3 ppl watching this post and its 2:59 wtf? smh

Answer 17

There's nothing else on the tv :P

Answer 18

lol same

Answer 19

yt is ded right now XD

Answer 20

type faster

Answer 21

`Faster`

Answer 22

._.

Answer 23

gn guys!

Answer 24

\[e^{\int\limits_{}^{} }\]haven't learned this in class hmm im confused
and this question is on the review sheet Q_Q

Answer 25

hmm :d

Answer 26

well, i'll just skip this one...i have a test today in few hours and i already have too much to cover

Answer 27

ahh good idea :D

Answer 28

\[\large\rm \int\limits_{ }^{} te^t dt = te^t- \int\limits_{}^{} e^tdt\]
dv= e^t dt , v=e^t
u=t
du = dt \[\large\rm t^2 Q= te^t-e^t+Chocolates\]

Answer 29

and then divide by t^2 that was easy @_@

Answer 30

Thanks