Question

y''+3y'+7y=cost , y(0)=0, y'(0)=2

Answer 1

actually, whats the question..?

Answer 2

the question is y''+3y'+7y=cost , y(0)=0, y'(0)=2

Answer 3

It is a second order differential equation. I don't remember the general solution

Answer 4

this is a statement @marielen123

Answer 5

It is e^(something)

Answer 6

laplace transforme

Answer 7

sorry I can't recall things let some experts help you @Hero @ganeshie8

Answer 8

step 1:
this is

\[s^2Y(s) - sy(0) - y'(0) + 3[sY(s) - y(0)] + 7Y(s) = \frac{ s }{ s^2 + 1 }\]

Answer 9

ok, I see

Answer 10

\[Y(s)[s^2 + 3s + 7] -s(0) - 2 - 3(0) = \frac{ s }{ s^2 + 1 }\]\[Y(s) = \frac{ 1 }{ s^2 + 3s + 7 } + 2 + \frac{ s }{ s^2 + 1 } \]

this is a pretty long one so i'll pause here to wonder how much more of it i need to do.

you'll want to make 1 fraction on the right and find the inverse transform of that [Y(s)], to find y(t)

Answer 11

what formula is this?

Answer 12

Laplace Transform:\[F(s) = \int\limits_{0}^{\infty}e^{-st} f(t) dt\]

most transforms of most f(t) are known, so we usually replace them directly without doing the whole integral process. there are many reference tables to get these

Answer 13

yes @Euler271

Answer 14

my difficulty is the fraction

Answer 15

ill be asking you questions to see how many steps i can skip. lmk if you want me to go over a step.

\[Y(s) = \frac{ s^2 + 1 + 2(s^2+1)(s^2 + 3s + 7) + s^3 + 3s^2 + 7s}{ (s^2 + 1)(s^2 + 3s +7) } = \frac{ 2s^4+7s^3+20s^2+13s+15}{ (s^2 + 1)(s^2 + 3s +7) }\]
\[Y(s) = = \frac{ 2s^4+7s^3+20s^2+13s+15}{ (s^2 + 1)(s^2 + 3s +7) }\]

are you familiar with partial frations? to turn the product on the right in a sum of fractions?

Answer 16

where
\[Y(s) = \frac{ As + B }{ s^2 + 1 } + \frac{ Cs + D }{ s^2 + 3s + 7 }\]

and we want to solve for A, B, C and D for our particular numerator

Answer 17

ok, now I understand, my teatcher taught differently

Answer 18

i've seen really weird ways to decompose the fraction lol. i like this way.

glad i could help

lmk if you have more questions ^_^

Answer 19

thank yoy very much ^-^

Answer 20

you*

Answer 21

Except that won't quite work. You will be missing your \(s^{4}\) in the numerator.

Answer 22

yes, really

Answer 23

Reform, first: \(\dfrac{2s^{4} + 7s^{3} + 20s^{2} + 13s + 15}{(s^{2}+1)(s^{2}+3s+7)} = 2 + \dfrac{s^{3} + 4s^{2} + 7s + 1}{(s^{2}+1)(s^{2}+3s+7)}\).

Then it will work.

Answer 24

this is true.

sorry for missing that. i didn't expect the problem to include almost every single "trick"

Answer 25

Just about!! Good work to both of you for dragging through all that!