Question #125203
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
1
Expert's answer
2020-07-06T19:42:48-0400

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

Making auxiliary equation, m4+2m3+11m2+18m+18=0m^4 + 2m^3 + 11m^2 +18m + 18 = 0

Roots of the equation are m=1±i,±3im = -1\pm i, \pm3i

So equation will be y=ex(c1cosx+c2sinx)+c3cos3x+c4sin3xy = 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)=23y(0) = 2,y'(0)=3,y''(0)=-11,y'''(0)=-23

simply putting x=0 in equation, we obtain

2=c1+c32 = c_1 + c_3 (1)

differentiating y once and applying condition , we get

y=ex(c1cos(x)+c2sin(x))+ex(c1sin(x)+c2cos(x))3c3sin(3x)+3c4cos(3x)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=c1+c2+3c43 = -c_1 + c_2 + 3c_4 (2)

differentiating again and applying conditions

y=2c1exsin(x)2c2excos(x)9c3cos(3x)9c4sin(3x)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=2c29c3-11 = -2c_2-9c_3 (3)

differentiating again and applying conditions

y=2c1(exsin(x)+excos(x))2c2(excos(x)exsin(x))+27c3sin(3x)27c4cos(3x)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=2c1+2c227c4-23 = 2c_1+2c_2-27c_4 (4)

solving equation 1,2,3 and 4,

c1=c2=c3=c4=1c_1=c_2=c_3=c_4 =1

Then equation will be

y=ex(cosx+sinx)+cos3x+sin3xy = e^{-x} ( cosx + sinx) + cos3x+sin3x



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