Thread: find value in ODE

Hi , Please Can I have some help with this problem:

Find the values of D such that the solution of the equation is bounded:

y''-4y'= sin(x); y(0)=D, y'(0)=0.

I started by applying the laplace Transform to the eqn.
I found (if not mistakes)

L(y)= [sD/(s^2-4) ] + [ 1/(s^2+1)(s^2-4)]

After that I know that I need to focus on the first term with D but how to apply the condition for bounded??

Thank you for your help

B

Originally Posted by braddy

Hi , Please Can I have some help with this problem:

Find the values of D such that the solution of the equation is bounded:

y''-4y'= sin(x); y(0)=D, y'(0)=0.

I started by applying the laplace Transform to the eqn.
I found (if not mistakes)

L(y)= [sD/(s^2-4) ] + [ 1/(s^2+1)(s^2-4)]

After that I know that I need to focus on the first term with D but how to apply the condition for bounded??

Thank you for your help

B

I do not understand what "bounded" means. Anyways, let me solve the differencial equation maybe that would help.
I think LaPlace Transforms is a primitive way to solve this.
Just use the charachteristic equation.
You have,

In such a case, there is a theorem. "The solution to a non-homogenenous equation is the sum of the general solution to the homogenous equation and a specific solution to the non-homogenous solution."

First solve the homogenous equation:
thus, its charachteristic is,
thus, . Thus, the general solutions to this homogenous equation are,


Second, find a specific solution to the non-homogenous,
. Since, there is a sine on the right look for solutions of the form,
thus,
thus,
.
Substitute that, into the non-homogenous

Thus,

Thus,

Solving, we have

Now, by the theorem, all general solution to this non-homogenous equation are the sum of these two solutions. Which gives,

----------
Looking at the initial conditions: Evaluating this at zero we get D,

Its, derivative is when evaluated at zero is,

Thus, solving these two equations,



Thus, the unique functions that satisfies this differencial equation with its initial conditions are:


Hope this helps.

Using Laplace transform is simpler. The system is unstable, because it has a pole in the right halfplane (s=4, yes you have mistake in your Laplace transform). Thats why we can see in the PerfectHacker's solution.
Because system is unstable there is no D for which the solution is bounded.