Asymptotic Behavior of Solutions to Linear Equations

[Xyius]

That is the title to the problem I am stuck on. Here is what it says..

Consider the equation,

$$\frac{dy}{dx} + ay = Q(x)$$

where a is positive and Q(x) is continuous on $$[0,\infty ],$$

Show that the general solution to the above equation can be written as..
$$y(x) = y(x_{0})e^{-a(x-x_{0})}+e^{-ax}\int^{x}_{x_{0}}e^{at}Q(t)dt$$

Where $$x_{0}$$ is a non-negative constant
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This part is easy, I just differentiated y(x) and plugged it into the equation. My real struggle is in the next part.
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If $$|Q(x)|\leq k for x\geq x_{0}$$ where k and $$x_{0}$$ are non-negative constante, show that..

$$|y(x)|\leq |y(x_{0})|e^{-a(x-x_{0})}+\frac{k}{a}[1-e^{-a(x-x_{0})}], for x\geq x_{0}$$
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The only thing I can come up with at the moment is plugging in the expression for y(x) and using the triangle inequality. I don't know where it would lead me though..

 

[jvicens]

Thank you for posting your question. I am a scientist and I would be happy to help you with this problem.

To begin, let's start with the general solution that you have already derived:

y(x) = y(x_{0})e^{-a(x-x_{0})}+e^{-ax}\int^{x}_{x_{0}}e^{at}Q(t)dt

To show that this solution satisfies the given equation, we can simply substitute it into the equation and see if it holds true. Let's do that now:

\frac{dy}{dx} + ay = \frac{d}{dx}[y(x_{0})e^{-a(x-x_{0})}+e^{-ax}\int^{x}_{x_{0}}e^{at}Q(t)dt] + a[y(x_{0})e^{-a(x-x_{0})}+e^{-ax}\int^{x}_{x_{0}}e^{at}Q(t)dt]

= -ae^{-a(x-x_{0})}y(x_{0})+ae^{-ax}\int^{x}_{x_{0}}e^{at}Q(t)dt + e^{-a(x-x_{0})}y(x_{0}) + e^{-ax}Q(x)

= ae^{-ax}\int^{x}_{x_{0}}e^{at}Q(t)dt + e^{-ax}Q(x)

= Q(x)

Therefore, our general solution does satisfy the given equation.

Now, to prove the inequality |y(x)|\leq |y(x_{0})|e^{-a(x-x_{0})}+\frac{k}{a}[1-e^{-a(x-x_{0})}], we can use the triangle inequality as you mentioned. Let's start with the left side of the inequality:

|y(x)| = |y(x_{0})e^{-a(x-x_{0})}+e^{-ax}\int^{x}_{x_{0}}e^{at}Q(t)dt|

\leq |y(x_{0})e^{-a(x-x_{0})}|+|e^{-ax}\int^{x}_{x_{0}}e^{at}Q(t)dt|

= |y(x_{0})|e^{-a(x-x_{0})}+|e^{-ax}\int^{x}_{x_{0}}e^{at}Q(t)dt|

Now