y'''' + 2y'' + y = 3t + 4,

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

y''(0) = y'''(0)=1

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- Nov 17th 2008, 04:42 PMjustin6mathhelpInitial Value Problem
y'''' + 2y'' + y = 3t + 4,

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

y''(0) = y'''(0)=1 - Nov 17th 2008, 04:50 PMThePerfectHacker
The charachteristic equation is,

$\displaystyle k^4 + 2k^2 + 1 = 0 \implies (k^2 + 1)^2 = 0 \implies (k-i)^2(k+i)^2=0$

Because the roots have multiplicity it means the general solution is,

$\displaystyle y = c_1\sin (t) + c_2 \cos (t) + c_3 t\sin (t) + c_4 t\cos (t)$

However, this only solves the equation $\displaystyle y'''' + 2y'' + y = 0$.

To finish the solution look for a particular solution which happens to be (for example) $\displaystyle 3t+4$.

Thus, the full solution is, $\displaystyle y = c_1\sin (t) + c_2 \cos (t) + c_3 t\sin (t) + c_4 t\cos (t) + 3t+4$