0
$\begingroup$

For:

$y''+ay'+by=0$

Given one of the solutions:

$y(x)=e^x cos(x)$

Whats the value of $a+b$ ?

I need some help with this, i dont know where to start.

$\endgroup$
1
  • 1
    $\begingroup$ Try substituting $y(x) = e^x \cos (x)$ into $y'' +ay' +by$ and equating it to zero. This should get you started. $\endgroup$ Apr 10, 2014 at 2:02

2 Answers 2

1
$\begingroup$

Find the first derivative and the second derivative.

Try to 'construct' the differential equation by using combinations of $y'$ and $y''$ and constants. Remember that you should have only 'one copy' of the double derivative since the question provided that the coefficient of $y''$ is 1.

Once you've got the differential equation, read off the values of $a$ and $b$.

$\endgroup$
4
  • $\begingroup$ I have the derivatives, and i tried substituting them in the equation, but dont know what to do with that because i have two unknown values a and b $\endgroup$
    – Wyvern666
    Apr 10, 2014 at 2:16
  • $\begingroup$ Don't substitute them into the equation. Construct the differential equation yourself and compare the coefficients of your equation with their one. For example, $y=\sin x \Rightarrow y'= \cos x \Rightarrow y''= -\sin x = -y$ Hence, this equation satisfies the differential equation $y''+y=0$. Do a similar thing but with $y=e^x \cos x$. $\endgroup$
    – Trogdor
    Apr 10, 2014 at 2:21
  • $\begingroup$ Ok, let me try. $\endgroup$
    – Wyvern666
    Apr 10, 2014 at 2:28
  • 1
    $\begingroup$ Ok, i find that $a=-2$, and $b=2$. But, for more complex expressions there is not a more procedural method to do this? $\endgroup$
    – Wyvern666
    Apr 10, 2014 at 2:33
1
$\begingroup$

Well the very first place to start is by subing in the given solution $y(x)$.

This may lead you to a constraint on a and b.

Do have you a problem doing the differentiation?

It is not so clear from your question where exactly the problem is.

$\endgroup$
1
  • $\begingroup$ Yes, i have done what you say, but dont know what to do next. $\endgroup$
    – Wyvern666
    Apr 10, 2014 at 2:21

You must log in to answer this question.

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