I'm not sure on how to start this higher order diff eq.

Question

Solve $$1001y''' + 3.2y'' + \pi y' - \sqrt{4}y = 0$$ Initial Conditions:

$$y(0) = 0, y'(0) = 0, y''(0) = 0$$

I plugged it into the characteristic equation, but I don't know where to go from there...

$$1001r^3 + 3.2r^2 + \pi r - 2 = 0$$

Some help would be appreciated, thanks.

Answer 1

Case 1: Assuming the last square root applies only to 4 (thus the question becomes trivial). WolframAlpha provides the solution:

$$ y(x) = c_3 e^{0.426569 x} + c_1 e^{-1.81328 x} sin(1.18346 x) + c_2 e^{-1.81328 x} cos(1.18346 x) $$

Which comes from the characteristic equation you gave. You would have to find the root either numerically or with Cardano's Method after a change in variables. https://en.wikipedia.org/wiki/Cubic_function

The coefficients $c_1,c_2$ and $c_3$ can be found with the initial conditions, but no surprise they're zero.

Case 1: The last square root applies to $y$ as well, so your question becomes a little more convoluted. You can still make a change of variables $u^2=y$, and the equation becomes:

$$ 1×(d^3 u(t)^2)/(dt^3) + 3.2×(d^2 u(t)^2)/(dt^2) + π×(du(t)^2)/(dt) - 2 u(t) = 0 $$

$$ (6.4 u(t) + 6 u'(t)) u^(3)(t) = 2 u(t) - 2 π u(t) u'(t) - 6.4 u'(t)^2 - 2 u(t) u^(3)(t) $$

While I fail to find a straightforward solution for this, you can check that again, no surprise the initial conditions being zero means that $y'''(0)$ is also zero, and thus $y(t)=0$ is a valid solution to your initial value problem.

Answer 2

I think the point of this problem is that, by the uniqueness theorem, there is only one solution, and from the initial conditions, it is clear that $y(t)=0$ is that solution. I don't think you were supposed to mess with the characteristic equation at all.