# Bounding Solutions of First Order ODE

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?

• See paragraph "First Order equation with variable coefficients" in this document 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 $$\int_{t_0}^tMe^{-p_0(t-s)}\,ds.$$