# Solve Linear ODE with Discontinuous f(x)

1s1

## Homework Statement

$$\frac{dy}{dx}+y=\left\{\begin{matrix}1, \ 0\leq x< 1 \\ 0, \ x\ge1 \ \ \ \ \ \ \ \end{matrix}\right.$$

## The Attempt at a Solution

$$P(x)=1$$
Integrating factor ##=e^{x}##

For ##f(x)=1##:
$$\frac{d}{dx}[e^{x}y]=e^{x}$$
Integrating both sides:
$$e^{x}y=e^{x}+C$$
$$y=1+\frac{C}{e^{x}}, y(0)=1$$
$$C=0$$
$$\fbox{y=1}$$

For ##f(x)=0##:
$$\frac{d}{dx}[e^{x}y]=0$$
Integrating both sides:
$$e^{x}y=C$$
$$y=\frac{C}{e^{x}}, y(0)=1$$
$$C=1$$
##y=\frac{1}{e^{x}}## <-- This one is coming back incorrect

I have worked through this problem multiple times, but my WeBWorK assignment keeps saying the second solution, the one for ##f(x)=0## is incorrect. I can't figure out what I'm doing wrong?! Any help would be greatly appreciated!

Homework Helper

## Homework Statement

$$\frac{dy}{dx}+y=\left\{\begin{matrix}1, \ 0\leq x< 1 \\ 0, \ x\ge1 \ \ \ \ \ \ \ \end{matrix}\right.$$

## The Attempt at a Solution

$$P(x)=1$$
Integrating factor ##=e^{x}##

For ##f(x)=1##:
$$\frac{d}{dx}[e^{x}y]=e^{x}$$
Integrating both sides:
$$e^{x}y=e^{x}+C$$
$$y=1+\frac{C}{e^{x}}, y(0)=1$$
$$C=0$$
$$\fbox{y=1}$$

For ##f(x)=0##:
$$\frac{d}{dx}[e^{x}y]=0$$
Integrating both sides:
$$e^{x}y=C$$
$$y=\frac{C}{e^{x}}, y(0)=1$$
$$C=1$$
##y=\frac{1}{e^{x}}## <-- This one is coming back incorrect

I have worked through this problem multiple times, but my WeBWorK assignment keeps saying the second solution, the one for ##f(x)=0## is incorrect. I can't figure out what I'm doing wrong?! Any help would be greatly appreciated!

Look at this :

$$y=\frac{C}{e^{x}}, y(0)=1$$

You're claiming ##x=0## for the initial condition, but ##x## is defined to be greater than or equal to 1.

1s1
Good point! I suppose I would need a different initial condition for this function?

1s1
Figured it out ... thanks Zondrina!

Have to appeal to the definition of continuity at a point.

So, you can say that ##Ce^{-1}=1##
$$C=e$$
And the solution for ##x \geq 0##
$$y=\frac{e}{e^{x}}$$

Thanks!!

Homework Helper
Gold Member
Sounds like there must be some fresh Prof. very keen on integrating factors and his students are coming here this last day or so using them where they're not needed!

You have a couple of nearly the most simple ode's there are!

1s1
Ha, ha good point! I suppose simple separation of variables would be SO much simpler. By the way, I'm new to working with ODE's, is using integrating factors pretty typical or are there alternative methods that are more common? Guess I'll probably find out more as I learn more about diffEq.

Homework Helper
Ha, ha good point! I suppose simple separation of variables would be SO much simpler. By the way, I'm new to working with ODE's, is using integrating factors pretty typical or are there alternative methods that are more common? Guess I'll probably find out more as I learn more about diffEq.

There are lots of different methods for solving first order linear ODEs. Some first order linear ODEs have more than one method which can be applied to find the correct answer ( As was in this case actually. You could have separated variables or used an integrating factor ). You're probably going to learn more methods along the way, but integrating factors and variable separation are very prominent methods.

Homework Helper
Gold Member
I mean to say it is just two equations; the second equation is just dy/dx = -y → dy/y = -dx
→ d(ln y) = -dx
and the first is d(y - 1)/dx = - (y - 1) , dealt with in same way.