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If y = \[5e ^{-4t} - 3e ^{-2t}\] show that \[\frac{ d^2y }{ dt^2 } + 6\frac{ dy }{ dt } +8y = 0\]

Derive Y?

use the formula
D(e^u)= e^u D(u) lets see what will you get

So Y' = -4(5e^-4t) + 2 (-3e^-2t) ?

e^ax = ae^ax

De^ax = ae^ax yup

So my first step is correct?

ah +3

\[= -4(5e ^{-4t}) + 2(-3e ^{-2t})\]

?

sorry about that i dint solved it my self but im just guiding you to the right solution :D

So am i right so far?

yes correct :D

Ok, what should I do next? :)

now do the 2nd derivative

Y'' yeah?

yup but you can still simplify that one by multiplying the constant coefficients

-4 multipled by 5e ?

Why is the 5e on the first line a -?

If you have -3e^-2t and the -2 comes down, wont you have +2 multiplying by -3 = -6? :D

its -3(-2)=+6

Oh :D

but for +3(-2)=-6

Sorry, so now do the 2nd derivative of this?

yeah try that

Any ideas how to start it? Aren't we back where we started?

I need to get an a/b no?

if
y'=-20e^-4t +6e^-2t
what is y" ?

or for example if
y=-20e^-4t +6e^-2t
what is dy/dx?

I know i have to derive that.
But not sure how to start

Do you use the same rule again?

do the same thing as you did on the first one

ah

yup the same rule applies to it

80e^-4t - 12e^-2t

? Don't know what you mean

_________ +6(_______) +8(________) =? solve it and see if you will get zero

But Don't i have a t?

its like 6 dy/dt
6( -20 e^-4t -3e^-2t ) multiply it

But how can you say it's equal to zero if you have T and Y still in the equation?

6 dy/dt =6( -20 e^-4t -3e^-2t )= -120 e^-4t -18 e^-2t

you will add or subtract like terms like 100 e^ 7t + 50 e^ 7t = 150 e^ 7t

But T has no value, no given value, so I can't say for example 80^-4t = 0!?

I don't know T! :P

Using your method I have : -40e^-4t + 24e^-2t + 8y = 0

yup it has no value its like this 10 e^ 7t - 10 e^ 7t = 0 e^ 7t=0

-40e^-4t + 24e^-2t + 8y = 0

:D

d^2y/dt^2 +6 dy/dt + 8y =?
(80e^-4t - 12e^-2t) +6(-20e^-4t +6e^-2t) + 8( 5e^-4t -3e^-2t)=?