Linear first order differential equation

Posted February 28th, 2008 by Structure

Let

$y'(x)+a(x)y(x)=f(x)$

where a and f are for the moment continuous function on an real closed interval.
We use a well known property of the definite Riemann integral of a continuous function, namely
let

$F(x)=\int_{x_0}^xa(t)dt$

then

$F'(x)=a(x)$

So, for

$G(x)=e^{\int_{x_0}^xa(t)dt}$

we have

$G'(x)=a(x)e^{\int_{x_0}^xa(t)dt}$

We may multiply our equation by G(x) and write

$y'(x)e^{\int_{x_0}^xa(t)dt}+a(x)e^{\int_{x_0}^xa(t)dt}y(x)=f(x)e^{\int_{x_0}^xa(t)dt}$

$y'(x)G(x)+G'(x)y(x)=f(x)e^{\int_{x_0}^xa(t)dt}$

which can be written

$(y(x)G(x))'=f(x)e^{\int_{x_0}^xa(t)dt}$

Now we want to take a definite integral in both sides , so we change the name of variables.

$(y(t)G(t))'=f(t)e^{\int_{x_0}^ta(\tau)d\tau}$

Integrating on $[x_0,x]$ we have

$\int_{x_0}^x(y(t)G(t))'dt=\int_{x_0}^xf(t)e^{\int_{x_0}^ta(\tau)d\tau}dt$

$y(x)G(x)-y(x_0)G(x_0)=\int_{x_0}^xf(t)e^{\int_{x_0}^ta(\tau)d\tau}dt$

But $G(x_0)=1$ so

$y(x)G(x)=y(x_0)+\int_{x_0}^xf(t)e^{\int_{x_0}^ta(\tau)d\tau}dt$

$y(x)e^{\int_{x_0}^xa(\tau)dt}=<br /> y(x_0)+\int_{x_0}^xf(t)e^{\int_{x_0}^ta(\tau)d\tau}dt$

$y(x)=e^{-\int_{x_0}^xa(\tau)d\tau}y(x_0)+e^{-\int_{x_0}^xa(\tau)d\tau}\int_{x_0}^xf(t)e^{\int_{x_0}^ta(\tau)d\tau}dt=$

$=e^{-\int_{x_0}^xa(\tau)d\tau}y(x_0)+\int_{x_0}^xf(t)e^{-\int_{x_0}^xa(\tau)d\tau}e^{\int_{x_0}^ta(\tau)d\tau}dt=e^{-\int_{x_0}^xa(\tau)d\tau}y(x_0)+\int_{x_0}^xf(t)e^{-\int_{t}^xa(\tau)d\tau}dt$

We have finally

$y(x)=e^{-\int_{x_0}^xa(\tau)d\tau}y(x_0)+\int_{x_0}^xf(t)e^{-\int_{t}^xa(\tau)d\tau}dt$

This formula shows thet our solution is a sum of two functions.
There is a very important interpretation for each of these two functions.
First we note that if f(x)==0 we have just the solution of the homogeneous equation

$y_0(x)=e^{-\int_{x_0}^xa(\tau)d\tau}y(x_0)$

Let

$y_p(x)=\int_{x_0}^xf(t)e^{-\int_{t}^xa(\tau)d\tau}dt$

For $y(x_0)=0$ our $y(x)=y_p(x)$ is a special solution of the non-homogeneous equation which take the value zero in $x_0$
It is not worthless to verify that from well known Lagrange formula
for

$y_p(x)=\int_{x_0}^xf(t)e^{-\int_{t}^xa(\tau)d\tau}dt$

we have

$y'_p(x)=f(x)+\int_{x_0}^x-a(x)f(t)e^{-\int_{t}^xa(\tau)d\tau}dt=-a(x)y_p(x)+f(x)$

$y'_p(x)+a(x)y_p(x)=f(x)$

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Linear first order differential equation

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