Question

5. Find the general solution to y'''-y''+4y'-4y = 0

Answer 1

For any equation,

`a_ny^(n)+\dots+a_1y'+a_0y=0`

assume solution of a form,
`e^(yt)`

Which leads to,

`(e^(yt))'''-(e^(yt))''+4(e^(yt))'-4e^(yt)=0`

Simplify to,

`e^(yt)(y^3-y^2+4y-4)=0`

Then find solutions,

`\underline{y_1=1}, \underline{y_2=2i}, \underline{y_3=-2i}`

For non repeated real root y, we have a form of,

`y_1=c_1e^t`

Following up,

For two non repeated complex roots
`y_2\\eq y_3` where,

`y_2=a+bi`

and,

`y_3=a-bi`

the general solution has a form of,

`y=e^(at)(c_2\cos(bt)+c_3\sin(bt))`

Or in this case,

`y=e^0(c_2\cos(2t)+c_3\sin(2t))`

Now we just refine and get,

`\boxed{y=c_1e^t+c_2\cos(2t)+c_3\sin(2t)}`

Hope this helps.

r3t40