Initial Value Problem

**justin6mathhelp** · November 17th 2008, 04:42 PM

y'''' + 2y'' + y = 3t + 4,
y(0)= y'(0) = 0,
y''(0) = y'''(0)=1

**ThePerfectHacker** · November 17th 2008, 04:50 PM

Originally Posted by justin6mathhelp

y'''' + 2y'' + y = 3t + 4,
y(0)= y'(0) = 0,
y''(0) = y'''(0)=1

The charachteristic equation is,
$k^4 + 2k^2 + 1 = 0 \implies (k^2 + 1)^2 = 0 \implies (k-i)^2(k+i)^2=0$
Because the roots have multiplicity it means the general solution is,
$y = c_1\sin (t) + c_2 \cos (t) + c_3 t\sin (t) + c_4 t\cos (t)$
However, this only solves the equation $y'''' + 2y'' + y = 0$ .

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

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

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