0
$\begingroup$

Given the Initial value problem: $y' + p(t)y = g(t)$ and $y(t_0) = y_0$. Show the solution can be written in the form:

$$y= y_0 \exp(-\int_{t_0}^{t}p(s)ds) + \int_{t_0}^{t}\exp(-\int_{s}^{t}p(r)dr)g(s)ds$$

Then, assuming that $0 < p_o \leq p(t)$ for all $t_0 \leq t$ and that $g(t)$ is bounded for $t_0 \leq t$, show that the initial value problem is bounded for $t_o \leq t$.

The first part comes from the integration factor for solving linear first order ODE. We are given that

$$0 < p_0 \leq p(t) \rightarrow0 > -\int p_0dt \geq -\int p(t) dt$$

When raised to the power of $e:$ $$I = e^{-\int p(t) dt} < 1$$

If $|g(t)| < M$ then in a similar fashion we can bound $$\int I g(t)dt < \int Mdt$$

And since I is bounded in this $t$ interval so will $g(t)$, and hence, the entire original expression is bounded.

Is this a correct justification?

$\endgroup$
1
  • $\begingroup$ See paragraph "First Order equation with variable coefficients" in this document $\endgroup$
    – Jean Marie
    Oct 3 at 20:57
1
$\begingroup$

No, obviously $$ \int_{t_0}^tM\,ds=M\,(t-t_0) $$ is not bounded. You might have greater success with removing less of the terms as in $$ \int_{t_0}^tMe^{-p_0(t-s)}\,ds. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.