Question

t^2(dx/dt)+3tx=t^4lnt+1, x(1)=0
solve initial value problem
this is what I got for standard form
dx/dt+3tx/t^2=t^4lnt/t^2+1/t2
final standard form
dx/dt+3x/t=t^2lnt+1/t^2.......not sure if its right

Answer 1

my algebra is HORRIBLE :(

Answer 2

Yah looks good so far :)

Were you able to find an integrating factor in order to solve for the general solution?

Answer 3

no I stopped there because I wasn't sure if I divided out the t^2 right...ill continue it now though

Answer 4

t^3 is what I got for the I.F.

Answer 5

ya me too, i think you're on the right track.

Answer 6

ok getting a little tough from here...not sure what to do

Answer 7

So you multiply through by your I.F. which gives us,\[\large t^3x'+3t^2x=t^5 \ln t+t\]

Don't forget to multiply BOTH sides by t^3 :)
Then our left side will simplify down a bit,\[\large (t^3x)'=t^5 \ln t+t\]Did you get that far?

Answer 8

sure did....glad I did that part right lol

Answer 9

give me a sec to try and get the next part

Answer 10

after integrating I got

3xt^2=(t^6/6)lnt+c

Answer 11

hmm

Answer 12

the left side should simply be giving us xt^3, dropping the brackets that tell us to take the derivative.

I'm a lil worried about your right side.. should have needed to apply Integration by Parts over there, giving you another term or two.

Answer 13

this is what I tried integrating
\[\int\limits_{?}^{?} d/dt (3xt^2)=\int\limits_{?}^{?}t^5lnt+t\]

Answer 14

oh ok I didn't use integration by parts :(

Answer 15

We had this, didn't we?
\[\large \int\limits (t^3x)'=\int\limits t^5 \ln t+t\;dt\]

Answer 16

my left side was
\[\int\limits t^3x'+3xt^2\]
guess I should've left out the 3xt^2 huh

Answer 17

i tried multiplying the IF which was t^3 with (3x/t) to give me 3xt^2

Answer 18

When you multiplied the left side by your integrating factor, you are correct, it gave you,\[\large t^3x'+3t^2x\]Which simplifies down to,\[\large (t^3x)'\]Right? :o

Answer 19

Product Rule in reverse :o

Answer 20

is that what happens at that point? reverse product rule? no wonder im having trouble

Answer 21

we multiply through by our integrating factor, and it gives us something of the form,\[\large uv'+u'v\]And yes, then we simplify it down using the product rule in reverse :)\[\large (uv)'\]


In our case, see how the 3t^2 is our (t^3)' ?

Answer 22

maybe that's confusing the way i said that lol

Answer 23

no that was helpful thank you

Answer 24

What we really have on the left side is,\[\large (t^3)x'+(t^3)'x\]

Answer 25

kk c:

Answer 26

well i attemped it and looked at the answer. IDK this is a tough one
what part would i have to do integration by part for?

Answer 27

The first term on the right requires Parts.\[\large \int\limits t^5 \ln t \;dt\]

Let \(\large u=\ln t\) and see if you can get it to work out :D

Answer 28

\(\large dv=t^5 dt\)

Answer 29

du i got t^5/6 and v is t^6/6

Answer 30

well u alone would be 1/t dt

Answer 31

du alone? ya

Answer 32

do i have to use u * v - integral vdu?

Answer 33

ya sounds right.

Answer 34

ok what i got was lnt (t^6/6)-(t^7/42) (1/t)

Answer 35

Woops, small boo boo on the second term.
You should have cancelled out those t's BEFORE integrating that second part.

So your v du gave you,\[\large \frac{1}{6}\int\limits t^6\frac{1}{t}dt\]Something like that, right?
So it would simplify to this before performing the actual integration.\[\large \frac{1}{6}\int\limits t^5 dt\]

Answer 36

yes that looks better

Answer 37

It'll just affect the fraction coefficient on that second term.

Answer 38

so its -2t^6/42 now?

Answer 39

oops make that 7t^6/42

Answer 40

lnt(7t^6/42)-(2t^6/42)

Answer 41

though i don't think that its 42 because the answer has 36 in the denominator

Answer 42

\[\large \frac{1}{6}\int\limits\limits t^5 dt\qquad=\qquad \frac{1}{6}\left(\frac{1}{5+1}t^{5+1}\right)\qquad=\qquad\frac{1}{36}t^6\]

Answer 43

I'm not sure where you keep getting this 7 from, heh

Answer 44

guess my v was wrong. i was using t^6/6 for v

Answer 45

oh but then the 1/t makes it t^5

Answer 46

ya :3

Answer 47

:/ geez my bad lol

Answer 48

lnt(1/36)?????

Answer 49

MMk let's just start over with this part -_- you're getting confused lol

Answer 50

sorry :(

Answer 51

\[\large \int\limits\limits \color{royalblue}{\ln t} \cdot \color{orangered}{t^5 \;dt}\]\[\large \color{royalblue}{u=\ln t}\qquad\qquad\qquad \color{orangered}{dv=t^5dt}\]\[\large \color{royalblue}{du=\frac{1}{t}dt}\qquad\qquad\quad \color{orangered}{v=\frac{1}{6}t^6}\]

So our integral gives us,\[\large \color{orangered}{\frac{1}{6}t^6}\color{royalblue}{\ln t}-\int\limits \left(\color{orangered}{\frac16t^6}\right)\color{royalblue}{\frac1t\;dt}\]Which before integrating again, simplifies to,\[\large \color{orangered}{\frac{1}{6}t^6}\color{royalblue}{\ln t}-\frac{1}{6}\int\limits t^5\;dt\]Which gives us,\[\large \frac{1}{6}t^6 \ln t-\frac{1}{36}t^6\]

Answer 52

For that portion.. blah i guess i didn't need to do all that color.. took me a bit too long lol

Answer 53

\[\large \int\limits\limits (t^3x)'=\int\limits\limits t^5 \ln t+t\;dt\]We can rewrite this as,\[\large \int\limits\limits (t^3x)'=\int\limits\limits t^5 \ln t \;dt+\int\limits t\;dt\]
And so far we've figured out this much,\[\large t^3x=\frac{1}{6}t^6 \ln t-\frac{1}{36}t^6+\int\limits t\;dt\]

Answer 54

ok i thought that we had to change 1/6 t^6 to work with the -1/36 t^6

Answer 55

so now we just divide everything by t^3 right?

Answer 56

well remember, we didn't integrate the +t on the right yet :o

Answer 57

oh yeah but that is just 1 so it could just be +C

Answer 58

woops, you accidentally took it's derivative instead of integrating.

Answer 59

:/ what a noob lol ok so t^2/2+C

Answer 60

heh

Answer 61

\[\large t^3x=\frac{1}{6}t^6 \ln t-\frac{1}{36}t^6+\frac{1}{2}t^2+C\]

Ok looks good so far

Answer 62

I really need to start writing the fractions as 1/6 t^6 instead of t^6/6 cause i forget about that 1/6 or fraction to take out of the integration

Answer 63

yah i like the fractions in front c: whatever works best for you though.

Answer 64

but from what we have now i just divide the t^3 out of the x and that should be the soln correct?

Answer 65

well actually i forgot about our initial condition. had to bring my laptop to the charger and left my work behind lol but ill scroll to the top

Answer 66

XD

Answer 67

Use the home and end keys if you have those on your keyboard :D nice quick way to get from bottom to top

Answer 68

cause i can imagine you're probably using a touchpad to scroll all the way up -_- oh boy lol

Answer 69

is c=-1?

Answer 70

hmmm

Answer 71

or 17/36 according to the ans i think lol

Answer 72

\[\large t^3x=\frac{1}{6}t^6 \ln t-\frac{1}{36}t^6+\frac{1}{2}t^2+C\]Plugging in our initial value,\[\large 1\cdot 0=\frac{1}{6}\cdot 1\cdot \ln 1-\frac{1}{36}\cdot1^6+\frac{1}{2}\cdot1^2+C\]Hmm I'm coming up with something different, your equation look the same?

Answer 73

ya the answer key is helpful heh.

Answer 74

well the final soln is
\[x=(t^3/6)lnt-t^3/36+1/2t-17/36t^3\]

Answer 75

Did you plug your 0 and 1 in correctly?
You shouldn't get -1 for C
D:

They should be plugged in as I did up above.
If your brain hurting too much at this point I understand XD lol

Answer 76

yeah i did do it wrong initially but i figured that it was because ln(0) doesn't work

Answer 77

i just wasn't sure about dividing out the t^3 before using our initial condition

Answer 78

Yah, before or after :D whichever you prefer.

Answer 79

yes its correct! thank you so much for bearing with me. this was def the hardest problem he assigned. I can't thank you enough for all your help!!!!!!

Answer 80

no probs \c:/ try to get a little more practice with integrating factor.
that product rule in reverse thing can be tricky to get used to.

Answer 81

yeah i know i need to review a lot. is there anything i can do to help you on the site besides giving you the best response?

Answer 82

aw thats sweet XD lol

naw i just like helping people :D
not looking for anything heh

Answer 83

just wrote a testimonial and became a fan! thanks again and have a good night

Answer 84

you too \c: