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?

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

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 $$ \int_{t_0}^tMe^{-p_0(t-s)}\,ds. $$


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