Question

By using integration by parts. Solve the solution for the following DE.
y'=(t^2-5)cos2t

Answer 1

Well, do it? What's stopping you? You \(may\) wish to break it up: \(t^{2}\cos(2t) - 5\cos(2t)\).

Answer 2

whats stopping me is that I need help on it

Answer 3

hahaha!! just take integral both sides

Answer 4

after taking the integral is it \[\frac{ t ^{3} }{ 3 }2\sin(2t)-5tsin(2t)(2)\]

Answer 5

Nope

Answer 6

http://www.wolframalpha.com/input/?i=int+%28t^2-5%29cos%282t%29+dt

Answer 7

Can someone show me what I did wrong?

Answer 8

@Loser66

Answer 9

where did you stuck?

Answer 10

in the integral

Answer 11

the whole integral? what is you start point?

Answer 12

integration by part! what did you do?

Answer 13

where did you come with this ?\(\frac{ t ^{3} }{ 3 }2\sin(2t)-5tsin(2t)(2)\)

Answer 14

you are abusing the rules of integration with that
anti derivatives

Answer 15

Do you know why?

Answer 16

from the equation tkhunny told me

Answer 17

yes but that's not how you integrate product of two function!!!
that rule work with one function standing by it self
or a function multiplied with a scalar
but here you have an operation that multiplies two functions
that's why i said you abused the rules!

Answer 18

oh i needed to use integration by parts

Answer 19

Recall that \(d/dx(f.g)(x)=d/x(f(x))g(x)+f(x)d/dx(g(x))\)

Answer 20

yes! by part. and the question is asking you to do so

Answer 21

im not really good with trig

Answer 22

well then you can't really solve this can you?
trig is important for calc!

Answer 23

set you integration by part first?

Answer 24

your*

Answer 25

if it helps remember that \(cos(2x)=cos^2x-sin^2x=2cos^2x-1\)

Answer 26

u=t^2
du=2t
dv=cos 2t
v=2sin(2t)
for the first integral

Answer 27

oh v=2cos^2x-1

Answer 28

why is v=2sin(2t)
?

Answer 29

No it is not
i'm just telling you that could be of a help

Answer 30

is it 2cosxsinx?

Answer 31

why? do you know your anti derivative
or derivatives of trig functions
at least cosine and sine?

Answer 32

yes the antiderivative of cos is sin

Answer 33

what's the problem here v=2sin(2t)

Answer 34

yes that correct! however you are dealing with a composite function
here correct? cos(2t)

Answer 35

dv=cos(ax) ====> v=sin(ax)/a

Answer 36

that's Chain rule there yes?

Answer 37

if v=2sint(2t) ===> dv=4cos(2t) which is incorrect, yes?

Answer 38

So what should you do

Answer 39

oh is sin(2t)/2

Answer 40

yes!

Answer 41

Now try working out the first one once you finished go back to the second part

Answer 42

so i have to take the integral for sin(2t)/2du

Answer 43

what is integral by parts ?

Answer 44

int (vdu)=uv-int(udv)

Answer 45

Actually you are looking for int(udv)
so int(udv)=uv-int(vdu) in your case

Answer 46

Do you got it?
u=t^2
dv=cos(2t) that's what you did, yes?

Answer 47

yes it would be \[\frac{ t ^{2}\sin(2t) }{ 2}-\int\limits \frac{ \sin(2t) }{ 2 }(2t)\]

Answer 48

hmm no! look carefully

Answer 49

oh sorry! that's correct

Answer 50

proceed...

Answer 51

so i take the antiderivative of the integral?

Answer 52

you have to repeat the integral by part! this is a problem that you need to work okay
you have to push it so you can solve this!

Answer 53

first cancel 2 you have int(tsin(2t))dt

Answer 54

you could have used u sub, but you need to learn this by part first

Answer 55

Are you working this or not! don't just look at it!
paper and pen okay hehehe

Answer 56

by the way did you take calc I, II?

Answer 57

hey are you working on it?