Find the value(s) of A such that the solution of the initial-value problem
y'' - 4y = sinx ; y(0)=A, y'(0)=0 is bounded
Answer is A=-1/10
Why is this the solution and how did they come up with it. I am pretty comfortable with laplace transforms and does it have to do with this.
Also
Find the value of A such that the solution of the initial-value problem
y' - 3y = 2e^(-2x) ; y(0)=A
has limit 0 as x goes to infinity.
Answer is A=-2/5
What's the differences in solving these two problems and what kind of solution do i use to solve them?
So the first one I use Laplace transform and get
(s^2)Y(s) - As - 4Y(s) = 1/((s^2)+1)
So You get 1/(((s^2)+1)((s^2)-4)) + As/((s^2)-4)
Separating the first part by partial fraction expansion
-1/(5 (s^2+1))-1/(20 (s+2))+1/(20 (s-2))
So I would have
-1/(5 (s^2+1))-1/(20 (s+2))+1/(20 (s-2)) + As/((s^2)-4) = Y(s)
(-1/20)e^-2x + 1/20e^2x + (-1/5)sinx + Acosh2x = f(x)
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Second Problem
sY(s) - A - 3Y(s) = 2/(s+2)
So
Y(s)= 2/((s+2)(s-3)) + A/(s-3)
Partial Fractions First Part
2/(5 (s-3))-2/(5 (s+2))
All together
2/(5 (s-3))-2/(5 (s+2)) + A/(s-3)=Y(s)
So
(2/5)e^3x - (2/5)e^-2x + Ae^3x = f(x)
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What do I do after this?