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I want to make sure I'm solving these differential equations correct. Is this right?

  1. $$y" - 2y' - 3y = 0$$ with initial conditions: $y(0) = 2$ and $y'(0) = 2$

so the auxiliary equation:

$$r^2 - 2r -3 = 0$$ $$(r + 1)(r-3) = 0$$ so $r = -1, 3$

so generally: $$y = c_1e^{-x} + c_2e^{3x}$$

if $y(0) = 2 = c_1 + c_2$ and $y'(0) = 2 = -c_1 + 3c_2$ then $c_2 = 1$ and $c_1 = 1$

so the solution given the initial conditions is:

$$y = e^{-x} + e^{3x}$$

Is that right?

  1. I'm a bit stuck on this one:

$$3y" + 4y' - 3y = 0$$

the auxiliary equation is:

$$3r^2 + 4r -3 = 0$$

The roots can't be found by factoring easily so I'll use the quadratic:

$$r = \frac{-4 \pm \sqrt{16 - 4(3)(-3)}}{6}$$

$$r = \frac{-4 \pm \sqrt{16 + 36}}{6}$$

$$r = \frac{-2}{3} \pm \frac{2 \sqrt{13}}{3}$$

so is the solution just:

$$y = c_1e^{r_1x} + c_2e^{r_2x}$$

where r1 and r2 are just the two roots found by solving the quadratic?

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The first solution is indeed correct. There's a slight error in the second, purely arithmetical:

$$3r^2 +4r - 3 = 0 \implies r = \frac{-4 \pm \sqrt{52}}{6} = \frac{-4 \pm 2\sqrt{13}}{6} = \frac{-2 \pm \sqrt{13}}{3}$$

When factoring out $\sqrt{52} = \sqrt{4} \cdot \sqrt{13}$, it seems you left it as just $4$ (as I did before this edit) instead of $2$. Thus your $r$'s should have $\sqrt{13}$, not double it.

Purely arithmetical and an easy oversight. Like I said, I did it too. Your overall idea is correct however for the second one.


Other than this, if I had to nitpick, if this is for, say, a school assignment, I would include how you solved for the $c_1,c_2$ on the assignment. Your values are right, it's largely a matter of making it easier to follow for whoever's reading. Might be just a personal thing though.

Also, note that you can always check these by plugging in the $y$ you get into the differential equation and checking, and also ensuring that the initial conditions are indeed met if applicable. Always a good paranoia check.


Edit:

While far from infallible, Wolfram Alpha also proves to be a good paranoia check. For example, the solutions for the first problem and second problem can be found there. If for some reason the expressions were to differ significantly from yours, that would be a sign of your derivation being wrong.

In this case, that's how I noticed both you and I were wrong in solving the second problem, since I could not see where that pesky $2$ disappeared to. :p

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  • $\begingroup$ EDIT: I derped and made the same mistake as you! Will fix. $\endgroup$ – Eevee Trainer May 4 at 3:46
  • $\begingroup$ And fixed, with explanation to boot. Sorry for the trouble I might have initially caused. $\endgroup$ – Eevee Trainer May 4 at 3:51

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