Differentiation Question! :(

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Differentiation Question! :(

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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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

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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 -?
hmm do it like this 5e^-4t -3e^-2t =5e^-4t D(-4t) -3e^-2tD-2t) =5(-4)e^-4t -3(-2)e^-2t =-20e^-4t +6e^-2t
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
yeah correct :D now do some multiplication and addition or subtraction using the formula to prove the one you wrote down above the d^2y/dx^2
? Don't know what you mean
just plug in the one you just solve to the formula d^2y/dt^2 +6 dy/dt +8y =0 _________ +6(_______) +8(________) =0
_________ +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
Sorry this is taking so long, I got A in my last 3 engineering differentiation exams, but just not getting this one
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
just plug in the one you just solve to the formula d^2y/dt^2 +6 dy/dt + 8y =? I I I V V V <------- arrow down pointing it solve them (_____________)+ 6(_________) + 8(_______)= __________ 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)=?
d^2y/dt^2 + 6 dy/dt + 8y =? (80e^-4t - 12e^-2t) + 6(-20e^-4t +6e^-2t) + 8( 5e^-4t -3e^-2t)=? solve by multiplying first the 6 times the value of dy/dt also 8 times the value of y then add or subtract like terms

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