Answer to Question #125203 in Differential Equations for Sahil

Question

Solve the initial value problem
y
iv + 2y
000
+ 11y
00
+ 18y
0 + 18 = 0, y(0) = 2, y
0
(0) = 3, y
00(0) = −11, y
000(0) = −23

Answer 1

Given equation is $y'''' +2y'''+11y''+18y'+18y=0$

Making auxiliary equation, $m^4 + 2m^3 + 11m^2 +18m + 18 = 0$

Roots of the equation are $m = -1\pm i, \pm3i$

So equation will be $y = e^{-x} (c_1 cosx + c_2sinx) + c_3cos3x+c_4sin3x$

applying boundary conditions $y(0) = 2,y'(0)=3,y''(0)=-11,y'''(0)=-23$

simply putting x=0 in equation, we obtain

$2 = c_1 + c_3$ (1)

differentiating y once and applying condition , we get

$y' = -e^{-x}\left(c_1\cos \left(x\right)+c_2\sin \left(x\right)\right)+e^{-x}\left(-c_1\sin \left(x\right)+c_2\cos \left(x\right)\right)-3c_3\sin \left(3x\right)+3c_4\cos \left(3x\right)$

$3 = -c_1 + c_2 + 3c_4$ (2)

differentiating again and applying conditions

$y'' = 2c_1e^{-x}\sin \left(x\right)-2c_2e^{-x}\cos \left(x\right)-9c_3\cos \left(3x\right)-9c_4\sin \left(3x\right)$

$-11 = -2c_2-9c_3$ (3)

differentiating again and applying conditions

$y'''=2c_1\left(-e^{-x}\sin \left(x\right)+e^{-x}\cos \left(x\right)\right)-2c_2\left(-e^{-x}\cos \left(x\right)-e^{-x}\sin \left(x\right)\right)+27c_3\sin \left(3x\right)-27c_4\cos \left(3x\right)$

$-23 = 2c_1+2c_2-27c_4$ (4)

solving equation 1,2,3 and 4,

$c_1=c_2=c_3=c_4 =1$

Then equation will be

$y = e^{-x} ( cosx + sinx) + cos3x+sin3x$