Question

Differentiation Question =]

Answer 1

Verify that the function V = 3sin(2t) + 4cos(2t) is a solution to the differential equation:

\[\frac{ d^2V }{ dt^2 } + 4V = 0\]

Answer 2

so did you try to solve it?

Answer 3

Not sure how to start hba. Got a pretty good knowledge of differentials, but stumped!

Answer 4

diff V twice.

Answer 5

Then?

Answer 6

diff V twice+4(3sin(2t) + 4cos(2t))

Answer 7

if (x,y) = (3,4) is a solution to: x - 2y = -5, then plugging in x=3 and y=4 will produce a "true" result

Answer 8

determining of something is a "solution" means to plug it into the setup and see if its true ....

Answer 9

Ok, makes sense..

Answer 10

HBA, what would the first derivative of it be?

Answer 11

@Asylum15 you know how to differentiate trigonometric functions don't you? I am not sure if Differential Equation is your major topic, but if it is, then it's absolutely imperative that you know your derivatives and integrals by heart.

Answer 12

Can I post what I have ?

Answer 13

sure :-)

Answer 14

3sin(2t) = product rule yeha?

Answer 15

Chain rule for this one, because it's a composition or conjuction of two functions, the sin(x) term where x=2t

Answer 16

Oh, and 4cos(2t) also chain rule..

Answer 17

f(x) = u(v(x)) ?

Answer 18

exactly

Answer 19

Could you give me an example of how 3sin(2t) turns out? Btw, I did differentiation 2 years ago, and resharpening it for engineering purposes (biomedical engineer)

Answer 20

alright then, always try to remember that the verbal version of the chain rule says "differentiation of the outside, with respect to the inside, and then differentiation of the inside".

So what that means is that you differentiate the outer function and keep the inner function the way it is, then you multiply the result by the derivative of the inner function.

So in your case:

\[y=3 \sin(2t)\]
turns out to be:
\[y'=3\cos(2t) * 2 = 6 \cos(2t)\]

Answer 21

Ah! Thus 4cos(2t) = -4sin(2t) x 2 = -8sin(2t)?

Answer 22

perfect.

Answer 23

Is the second derivative 2 more chain rules?

Answer 24

it works analog to what we did yes, you computed the first derivative now, just do it the same way as above to obtain the second derivative. Then plug that into your given problem.

Answer 25

Then add it to four times the equation?

Answer 26

Not quite, V denotes the function itself, thus the solution to your given Differential Equation as suggested in your problem.

V' denotes the first derivative, it's Leibniz Notation would be dV/dt, analog you can denote the second derivative as V'' and in Leibniz form by dV^2/(dt^2) so you plugin the second derivative there.

Answer 27

I understand! Thank you so much spacelimbus!

Answer 28

you're very welcome.