Use the laplace transform to solve the initial value problem: y''+y=1, y(0)=2 and y'(0)=0
1 answer:
Answer:
y(s) = sin t + 2 cost
Step-by-step explanation:
given,
y'' + y = 1, y(0) = 2 and y'(0) = 0
applying Laplace transformation both side
s²y(s) - s y(o)- y'(0) + y(s) = L{1}
s²y(s) - 2 s + y(s) = L{1}
y(s)(s² + 1 ) = L{1} + 2 s
y(s) = sin t + 2 cost
hence, the required solution of Laplace will be
y(s) = sin t + 2 cost
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Answer:
Step-by-step explanation:
Use the Difference of Cubes [(a³ - b³)(a² + 2ab + b²)] to figure this out:
Then, since the divisor\factor is in the form of <em>x - c</em>, use what is called Synthetic Division. Remember, in this formula, −c gives you the OPPOSITE terms of what they really are, so do not forget it. Anyway, here is how it is done:
−2| 1 0 0 8
↓ −2 4 −8
___________
1 −2 4 0 →
You start by placing the <em>c</em> in the top left corner, then list all the coefficients of your dividend [x³ + 8]. You bring down the original term closest to <em>c</em> then begin your multiplication. Now depending on what symbol your result is tells you whether the next step is to subtract or add, then you continue this process starting with multiplication all the way up until you reach the end. Now, when the last term is 0, that means you have no remainder. Finally, your quotient is one degree less than your dividend, so that 1 in your quotient can be an x², the −2x follows right behind it, then 4, giving you the other factor of , attaching this to the first two factors you started out your work on:
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