0
$\begingroup$

If $y=e^{x+a}+be^x$ which one will be the 'correct' ODE to which it will be a solution to considering $a$, $b$ to be the parameters? $$y''-y=0 \text{ or } y'-y=0$$

In other words can $e^a+b$ chosen as a single parameter if nothing has been mentioned?

$\endgroup$
4
  • $\begingroup$ I'm assuming that second equation should have a $+$ in it rather than the first $=$ sign. If so the first one yields complex solutions while the second does not $\endgroup$
    – Triatticus
    Commented Jan 4, 2018 at 5:17
  • $\begingroup$ I think OP meant $-$ in both ODEs. $\endgroup$
    – jaslibra
    Commented Jan 4, 2018 at 5:21
  • $\begingroup$ Did you try anything...? $\endgroup$
    – Myridium
    Commented Jan 4, 2018 at 7:34
  • $\begingroup$ I have chosen $a$ and $b$ to be two different parameters and differentiate it twice to obtain an ODE of the form $y''=y=0$. Why cannot I choose $a$ and $b$ to be two different parameters?. $\endgroup$
    – Alexander
    Commented Jan 4, 2018 at 9:51

3 Answers 3

1
$\begingroup$

by saying $e^a+b=C$ you get $y=Ce^x$. this can be a solution for both $y''-y=0$ and $y'-y=0$, but let's solve those:

  • $y''-y=0:$ assuming $y=e^{\gamma x}\implies\left(e^{\gamma x}\right)''-e^{\gamma x}=e^{\gamma x}(\gamma^2-1)=0\implies\gamma=\pm1$ hence the general solution is $y=c_1e^x+c_2e^{-x}$
  • $y'-y=0:$ we can rearrange it into $\frac1yy'=1\implies\int\frac1ydy=\int1dx$ hence $\ln y=x+c\implies y=e^{x+c}=Ce^x$

so we can see that $y=Ce^x,C=e^a+b$ is a general solution for $y'-y=0$, even though it is also solution for $y''-y=0$.

The word correct is not really the right word because we can plug in $y=Ce^x,C=e^a+b$ to both of them and see that it is solution for both of them, the different is that any ODE in the form of $y'-y=0$ can be expressed like this and not all ODE in the form of $y''-y=0$ can so $y'-y=0$ is more 'correct'

$\endgroup$
0
$\begingroup$

The only way that this question makes sense is if both of them have minus signs.

The $y = ce^x$ with $c = e^a + b$ you gave is a solution to both of the differential equations and so both are correct.

$\endgroup$
1
  • $\begingroup$ However, it's not the general solution of $y''-y=0$. $\endgroup$ Commented Jan 4, 2018 at 5:26
0
$\begingroup$

The correct answer $y'-y=0$.Note that $ y=e^{x+a}+be^x $ is $y=e^x(e^a+b)=ke^x.$ Solutions to the equation $y''-y=0$ are linear combinations of $e^x$ and $e^{-x}.$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .