Question

Help me out please.
\(x'+\dfrac{2}{(150+0.5t)}x =7.5\)
I got trouble with \(e^{\int p(t) dt}\)

Answer 1

\(\int p(t) dt = \int \dfrac{2}{150+0.5t}dt\)
Case 1: \(dfrac{2}{150+0.5t}=\dfrac{4}{300+t}\)
Hence when taking integral, I got \(4 ln(300+t)\), when plugging in, \(\mu= (300+t)^4\)

Answer 2

Case2, If I let it as it is,
Let u = (150+0.5t) then 2du = dt
therefore \(\int \dfrac{4du}{u}=4ln u= 4ln(150+0.5t)\)
Plugging in, I got \(\mu= (150+0.5t)^4\)
OHOHOH>>>> NO.
What is wrong?

Answer 3

hint: everyone forgets about the constant of integration...
Is not ln(ax) = ln(x)+ln(a)?

Answer 4

I don't get what you mean. We don't have ln (ax), we have a*ln x

Answer 5

ok try this problem. Is integral of ln(x) then same as ln(ax) for some constant a?

Answer 6

\[\int\limits \ln x dx = \frac{ 1 }{ x } + C \]
\[\int\limits \ln(ax)dx = a(\frac{ 1 }{ ax }) +C = \frac{ 1 }{ x } + C\]



So there's no difference really)

Answer 7

@ganeshie8 help me please.

Answer 8

Yeah there can be infinitely many integrating factors, just need one to use the product rule in reverse.

Answer 9

no, ganeshie8, the result must be the same so that we can get the correct solution.

Answer 10

All give you the correct solution

Answer 11

\((300+t)^4\neq (150+0.5t)^4\)

Answer 12

Who said they need to be equal ?

Answer 13

At the end, we put initial condition to solve for C.

Answer 14

Right, all integrating factors give you the same solution for IVP

Answer 15

And if they are different, I have 2 different C's . How can I know which one is correct one?

Answer 16

The particular value of C will change based on the general solution

Answer 17

IC x(0) =20
hence , the first one gives me C =-430 (300)^4
the second one gives me C =-430(150)^4
OHOH, no

Answer 18

And what are the general solutions ?

Answer 19

For the first one: \(x(t)=1.5(300+t)+\dfrac{C}{(300+t)^4}\)

Answer 20

The second one is \(x(t) =3(150+0.5t)+\dfrac{C}{(150+0.5t)^4}\)

Answer 21

Yeah the C can change so don't think C1 = C2. common mistake.

Answer 22

Is it hard to see that you get identical equations after you plugin the respective C's ?

Answer 23

Oh, I got you. MY GOD, how stupid I am.
Thanks a ton.

Answer 24

yes, yes, yes, lalala... hahaha... I am struggling with it more than 2 hours.

Answer 25

Haha nope. you're not stupid. lookup existence and uniqueness theorem. It's proof is hard but you might enjoy :)

Answer 26

I have final on next Monday, must be ok with all of the materials.

Answer 27

good luck with the exam !

Answer 28

Thank you :)