Question

Differential Equations Initial Value Problem
I worked it out, but I arrived at a different answer than is given in my text book.

\(ty\prime + (t+1)y=t\)
\(y(\ln(2))=1,~t>0\)

Answer 1

I arrived at
y = t + 1 - ln(2)

My books answer....
\(y=\dfrac{(t-1+2e^{-t})}{t},~t\ne0\)

Answer 2

You have a new DE and you didn't invite me? T.T

Answer 3

@terenzreignz :P

Answer 4

My way is clearly incorrect. I think I need to treat the t+1 term differently.
Any thoughts?

Answer 5

I say go for that integrating factor again... After dividing everything by t, then we get the integrating factor ro be \[\Large \mu = \int \frac{t+1}{t}dt\]

Answer 6

Darn it.... stupid stupid stupid stupid stupid Austin.

Answer 7

Whoops... I meant...\[\LARGE e^\mu\] as the integrating factor...

Answer 8

\(\huge{\mu(t)=te^t}\)

Answer 9

Look ok?

Answer 10

That's the integrating factor itself, so... okay, great :D
Multiply everything with it... you get
\[\Large te^t y' + (t+1)e^ty = te^t\]

Answer 11

That's the integrating factor itself, so... okay, great :D
Multiply everything with it... you get
\[\Large te^t y' + (t+1)e^ty = te^t\]

Answer 12

\(\dfrac{(t+1)}{t}\)
Right?

Answer 13

What do you mean?
THIS term?
\[\Large te^t y' +\color{red}{ (t+1)e^ty} = te^t\]

Answer 14

Yes.

Answer 15

the standard equation for a linear differential equation is
y' +Py = Q

the first step towards solving the differential equation lies in identifying weather the given equation is in standard form or not. This can be achieved by comparing the given equation with the standard form.

By doing so, we find that the given eq. isn't in the standard form
so we must do certain operations to bring it to the standard form
in this case
We divide the both sides with 't'
to obtain y' + (t+1)/t = 1

here, P=(t+1)/t
or P =t/t +1/t
or P= 1+1/t

IF = e^int(p dt)
IF =e^[int(1) + int(1/t )]
IF =e^[t+log t]
IF = e^t + e^log t
IF = e^t + t

the solution is given by
ye^int(p dx) = int(Q e^int(p dx))

Answer 16

Remember, we started with this:
\[\Large t y' + (t+1)y = t\]
divided everything by t
\[\Large y' + \frac{(t+1)}ty =1\]
Multiplied everything by \(\large te^t\), the integragint factor, arriving finally at:
\[\Large te^t y' + (t+1)e^ty = te^t\]
Questions? :P

Answer 17

Oh... duh Austin. I swear I don't function without caffeine. I feel like I just went full dipstick.

Answer 18

Okay, well, I don't need caffeine, but some mocha ice cream would be awsum ^_^
Anyway, left side is now
\[\Large \frac{d}{dt}[te^ty]= te^t\]
Let's have at it Austin :3

Answer 19

Yep, I have worked that out. And then I have arrived at C=2

Answer 20

\(te^ty=(t-1)e^t+2\)

Answer 21

Does that work out?

Answer 22

Oh, it IS right, then?

Answer 23

I believe so!

Answer 24

Then... ice cream time :D
More ice cream, that is ^_^

Answer 25

http://www.simplyrecipes.com/wp-content/uploads/2007/04/coffee-ice-cream.jpg?ea6e46

Answer 26

Lovely :D
Well, I'm off ^_^

Have fun XD
------------------------------
Terence out

Answer 27

Thanks for all of your help Terence!