A Question in 1st order ODE Hi I have a question when solving exercise from ODE by E.A.Coddington
page 42 Q8 as follows 
consider the equation $ y'+ay=b(x )$, where a is a constant, and b is a continuous function on $o \le x <\infty $, satisfying there $|b(x)| \le k $, where k is some positive number.
(a) Find the solution satisfying y(0)=0 
(b) If Re(a) $ \neq $ 0, Show that the solution satisfies $$|y(x)| \le \frac {k}{Re(a)} [1- e^{-(Re(a)x} ] $$
I am lost on how to prove part a as i have only one property about b. 
please any help appreciated 
Thanks
 A: For (a) – and this is standard textbook stuff, though it may be presented differently – if $u(x)=e^{ax}y(x)$ then you get $u'=e^{ax}(y'+ay)=e^{ax}b(x)$, leading to $u(x)=u(0)+\int_0^x e^{at}b(t)\,dt$. Substituting back, you get $$y(x)=y(0)e^{-ax}+\int_0^x e^{a(t-x)}b(t)\,dt=\int_0^x e^{a(t-x)}b(t)\,dt$$
(since $y(0)=0$).
For (b), the formula above gives
$$|y(x)|\le\int_0^x |e^{a(t-x)}b(t)|\,dt
  \le k\int_0^x|e^{a(t-x)}|\,dt.$$
Finally, for any complex $z$ use $|e^z|\le e^{\operatorname{Re}z}$ to estimate the integral. May I leave the last bit to you?
A: Here's my attack on this problem:
Starting with the given equation
$$y' + ay = b(x), \tag{1}$$
we follow the suggestion of Harald Hanche-Olsen's comment and set
$$u(x) = e^{ax}y(x). \tag{2}$$
It should be noted at this point that this is a well-known and standard approach to equations of the general form (1), and if you search the web for topics such as integrating factors and the variation of parameters method, you will quite likely find a panoply of references/citings to this technique.  In any event, making the substitution (2) we see that
$$u'(x) = ae^{ax}y(x) + e^{ax}y'(x) = e^{ax}(y'(x) + ay(x)), \tag{3}$$
leading to
$$u'(x) = e^{ax}(y'(x) + ay(x)) = e^{ax}b(x), \tag{4}$$
which may immediately integrated, yielding
$$u(x) - u(0) = \int_0^x e^{as}b(s)\, ds, \tag{5}$$
and since $u(0) = y(0) = 0$, $\text{(5)}$ becomes
$$u(x) = \int_0^x e^{as}b(s)\, ds, \tag{6}$$
or
$$e^{ax}y(x) = \int_0^x e^{as}b(s)\, ds, \tag{7}$$
or
$$y(x) = e^{-ax} \int_0^x e^{as}b(s) ds, \tag{8}$$
the sought solution for $y(x)$.  Here, two points are worthy of note:  first, this solution cannot be taken further without greater, specific information regarding $b(x)$; and second, the factor $e^{ax}$ is known as the integrating factor for (1), since its introduction via (2)-(4) allows the resulting equation to be integrated forthwith, as in (5).  In any event, having (8) in hand, we may proceed to estimate $\vert y(x) \vert$, as follows:  by (8),
$$\vert y(x) \vert = \vert e^{-ax} \int_0^x e^{as} b(s) ds \vert = \vert e^{-ax} \vert \vert \int_0^x e^{as}b(s) ds \vert \le \vert e^{-ax} \vert  \int_0^x \vert e^{as}\vert \vert b(s) \vert ds. \tag{9}$$
We look at the exponential factors occurring in (9):  we have, for any complex $a = Re(a) + i Im(a)$,
$$\vert e^{ax} \vert = \vert e^{Re(a) x} e^{i Im(a) x} \vert = \vert e^{Re(a) x} \vert \vert  e^{i Im(a) x} \vert = \vert e^{Re(a) x} \vert = e^{Re(a) x}, \tag{10}$$
since $\vert  e^{i Im(a) x} \vert = 1$ and $e^{Re(a) x} > 0$ for all real $x, Re(a), Im(a)$.  We thus have for the right-hand integral in (9), also using $\vert b(x) \vert \le k$:
$$\int_0^x \vert e^{as}\vert \vert b(s) \vert  \le k \int_0^x e^{Re(a) s} ds = (k / Re(a))[e^{Re(a) x} - 1]; \tag{11}$$
returning to (9), using (11),
$$\vert y(x) \vert \le \vert e^{-ax} \vert  \int_0^x \vert e^{as}\vert \vert b(s) \vert ds$
$\le e^{-Re (a) x} (k / Re(a))[e^{Re(a) x} - 1] = (k / Re(a))[1 - e^{-Re(a) x}], \tag{12}$$
which is the required estimate.
Hope this helps.  Happy New Year,
and as always,
Fiat Lux!!!
