# Math Help - find value in ODE

1. ## 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

2. 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,
$y''-4y'=\sin x$
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:
$y''-4y'=0$ thus, its charachteristic is,
$k^2-4k=0$ thus, $k=0,4$. Thus, the general solutions to this homogenous equation are,
$y=C_1e^{0x}+C_2e^{4x}=C_1+C_2e^{4x}$

Second, find a specific solution to the non-homogenous,
$y''-4y'=\sin x$. Since, there is a sine on the right look for solutions of the form,
$y=A\sin x+B\cos x$ thus,
$y'=A\cos x-B\sin x$ thus,
$y''=-A\sin x-B\cos x$.
Substitute that, into the non-homogenous
$
-A\sin x-B\cos x-4A\cos x +4B\sin x=\sin x
$

Thus,
$(-A+4B)\sin x+(-B-4A) \cos x=\sin x$
Thus,
$\left\{\begin{array}{c}-A+4B=1\\-B-4A=0\end{array}\right$
Solving, we have $(A,B)=(1/15,4/15)$

Now, by the theorem, all general solution to this non-homogenous equation are the sum of these two solutions. Which gives,
$y=C_1+C_2e^{4x}+\frac{1}{15}\sin x+\frac{4}{15}\cos x$
----------
Looking at the initial conditions: Evaluating this at zero we get D,
$C_1+C_2+\frac{4}{15}=D$
Its, derivative is when evaluated at zero is,
$4C_2+\frac{1}{15}=0$
Thus, solving these two equations,
$C_1=D-1/4$
$C_2=-1/60$

Thus, the unique functions that satisfies this differencial equation with its initial conditions are:
$y=D-\frac{1}{4}-\frac{1}{60}e^{4x}+\frac{1}{15}\cos x-\frac{4}{15}\sin x$

Hope this helps.

3. ## Simpler

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 $C \exp(4x)$ in the PerfectHacker's solution.
Because system is unstable there is no D for which the solution is bounded.
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