Thread: Find value so that initial value problem is bounded

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?

Originally Posted by bustacap09

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?

Since you're comfortable with Laplace Transforms, start by solving each DE using them. Show us what you have done then we'll help you with the analysis questions that follow regarding values of A.

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?