Question

3rd order linear homogeneous diff equation with constant coefficient

y'''=y ; y(0)=1, y'(0)=0, y''(0)=0

Answer 1

Just curious: are you allowed to use Laplace Transforms here? If so, this will be not too bad. If not, we're going to be in for quite the ride. XD

Answer 2

i dont recall him mentioning that method i only know that we rewrite \[\lambda ^ 3 - 1=0\]

Answer 3

and well solve for landa which i think it would be landa = 1 multiplicity 3
and making the answer y=Ax^2e^x+Bxe^x+ce^x
but how ever the answer is wrong

Answer 4

Ok, so that implies then that

\(\large (\lambda-1)(\lambda^2+\lambda+1)=0\implies \lambda-1=0\text{ or }\lambda^2+\lambda+1=0\).

Here, we get one real root and two complex roots. The real root is \(\large \lambda = 1\) and the complex roots are found using the quadratic formula:
\[\begin{aligned}\lambda^2+\lambda +1 = 0\implies \lambda &= \frac{-1\pm\sqrt{1-4}}{2}\\ &= -\frac{1}{2}\pm i\frac{\sqrt{3}}{2}\end{aligned}\] Therefore, the general solution is
\[\large y(t) = c_1e^t + e^{-t/2}\left(c_2\cos\left(\frac{\sqrt{3}}{2}t\right) + c_3\sin\left(\frac{\sqrt{3}}{2}t\right) \right)\] Now you need to differentiate this general equation 3 times and plug in your initial conditions. At this point, you can then solve a system of equations for the constants \(\large c_1,c_2,c_3\).

Can you take things from here?

Answer 5

Sorry, I mean you need to find \(y^{\prime}(t)\) and \(y^{\prime\prime}(t)\) (only need to differentiate twice, not three times) and then plug in the initial conditions and solve for the unknown constants.

Sorry about that! :-/

Answer 6

yea i got it from there thank you so much :)