Question

x^2-16

Answer 1

Simply solve this Differential equation.

Answer 2

Is it y^4 or y''''?

Answer 3

Ok, start by assuming that the solution will be of the form \(y = e^{\lambda t}\), and substitute it in:
\[y'''' = \lambda^4e^{\lambda t}\]
\[y''' = \lambda^3e^{\lambda t}\]
\[\lambda^4e^{\lambda t} - \lambda^3e^{\lambda t} = 0\]
\[e^{\lambda t} \left(\lambda^4- \lambda^3\right) = 0\]

Now solve for lambda and get 4 linearly independent functions, and add them together to get the general solution

Answer 4

It can be factored as:
\[e^{\lambda t}\left(\lambda^4 - \lambda^3\right) = 0\]
Now since \(e^{\lambda t}\) can't be zero, the zeros must come from the other part, ie;
\[\lambda^4 - \lambda^3 = 0\]
Now you could factor the lambda out
\[\lambda^3\left(\lambda - 1\right)=0\]

Answer 5

yes, but 0 has multiplicity 3 so you have to take that into consideration

Answer 6

Well basically, if \(\lambda\) is a root of multiplicity \(k\), then the solutions from that root are
\[e^{\lambda t},te^{\lambda t},\cdots,t^{k-1}e^{\lambda t}\]

So for your roots (since we have 4, we get 4 functions)
\(y(t) = e^t\) (root 1)
\(y(t) = 1\) (root 0)
\(y(t) = t\)
\(y(t) = t^2\)

so basically if you have repeated roots you just keep chucking a t in front of it until you have k functions where k is the time the root is repeated

Now according to the principle of superposition, the general solution is
\(y(t) = c_1e^t + c_2t^2 + c_3t + c_4\)

Now you would apply the initial conditions and find as much of these constants as possible

Answer 7

Well you initial conditions are:
\[y(0) = 0\\y'(0)=0\\y''(0)=0\\y'''(0) = 0\]

So before we get started on applying these, lets first find the derivatives:
\[
y(t) = c_1e^t + c_2t^2 + c_3t + c_4\\
y'(t) = c_1e^t + 2c_2t+c_3\\
y''(t) = c_1e^t + 2c_2\\
y'''(t) = c_1e^t
\]

So lets start by substituting in:
\[
y(0) = c_1 +c_4=0\\
y'(0) = c_1 + c_3=0\\
y''(0)=c_1+2c_2=0\\
y'''(0)=c_1=0\]

this gives the solutions:
\[c_1 = 0\\c_2 = 0\\c_3=0\\c_4 = 0\]
and makes the general solution:
\[y(t) = 0e^t + 0t^2 + 0t + 0 = 0\]

Answer 8

no problem :)