# Linear ordinary differential equation.

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1. Jun 4, 2017

### Buffu

1. The problem statement, all variables and given/known data

$\dfrac{dy}{dx} + y = f(x)$

$f(x) = \begin{cases} 2 \qquad x \in [0, 1) \\ 0 \qquad x \ge 1 \end{cases}$

$y(0) = 0$

2. Relevant equations

3. The attempt at a solution

Integrating factor is $e^x$

$e^x\dfrac{dy}{dx} + e^x y = e^x f(x)$

$\displaystyle ye^x = \int e^x f(x) dx = \begin{cases} \int 2e^x dx = 2e^x + C \qquad x \in [0, 1) \\ \int 0 dx = C^\prime \qquad x \ge 1 \end{cases}$

Solving for inital value,

$ye^x = \begin{cases} 2(e^x - 1) \qquad x \in [0, 1) \\ 0 \qquad x \ge 1 \end{cases}$

The answer is $ye^x = \begin{cases} 2(e^x - 1) \qquad x \in [0, 1) \\ 2(e - 1) \qquad x \ge 1 \end{cases}$
, which is way off.

I think the fault is in my answer this time but I can't point it :(.

2. Jun 4, 2017

From your first solution $y(1)=2(1-e^{-1} )$. (We're going to assume that $y$ is continuous at $x=1$). From the second solution $y \, e^x=C'$ . But $C'=y(1) e^1$. Thereby $y e^x=2(e-1)$ for $x>1$.

3. Jun 4, 2017

### epenguin

Finding in integration factor seems over-engineering. You could just as well treated as two differential equations, both very simple:

y' = -y and y' = - ( y - 2 ) .

The arbitrary constant for the solution of the second is in principle arbitrary, but it would be natural to treat it as given by a requirement of continuity as already mentioned.

4. Jun 4, 2017

One question/comment for @Buffu is how do you know that the homogeneous solution $y=A \, e^{-x}$ doesn't play a role in the solution? That is apparently an underlying assumption in the problem (that you ignore any homogeneous solution), but it is always good practice to consider the possibility of a homogeneous solution. $\\$ Editing: In the solution for $x>1$, the solution is in fact the homogeneous solution. Also, for $0<x<1$, the integrating factor method you used actually generated a homogeneous solution. $\\$ Let me offer an alternative solution to the differential equation above: For $0<x<1$, $y_p=2$ and the homogeneous solution is $y_h=A \, e^{-x}$, so that for $0<x<1$ , $y=2+A \, e^{-x}$ for some $A$ . $\\$ For $x>1$, $y_p=0$, and $y_h=B \, e^{-x}$, so that $y=B \, e^{-x}$ for some $B$. $\\$ The constants $A$ and $B$ are easily found.

Last edited: Jun 5, 2017
5. Jun 5, 2017

@Buffu I edited my comments in post #4, because I see your solution, using an integrating factor, actually resulted in generating the necessary homogeneous solution. There is, as @epenguin pointed out, a simple solution that doesn't require an integrating factor. Please see my edited post #4.

Last edited: Jun 5, 2017
6. Jun 5, 2017

### Buffu

Yes I see what mistake I made.
Thank you.

For the integrating factor, I did not see that DE here is a simple separable because I overlooked the condition on $f(x)$. I thought, let me reduce the equation first then I will care about $f(x)$.