Initial Value Problems
y ''' + 2y'' - 5y' - 6y = 0, y (0) = y' (0) = 0, y '' (0) = 1
Assume y = y(x) = erx
Then, the characteristic polynomial is :
r3 + 2r2 – 5r – 6 = 0
By factorization :(r+1)(r-2)(r+3) = 0
Roots are : r = -1, +2, -3
Thus, y(x) = C1e-x + C2e2x + X3e-3x
And y’(x) = -C1e-x + XXXeXX - 3CXe-3x
Y”(x) = CXe-x + XXXeXX + XX3e-3x
y (0) = y' (X) = X, y '' (0) = X,
XX,y(0) =X1eX + X2e0 + CXeX orC1 + C2 + X3 = 0 ........i
y' (0) =-C1eX + XXXeX - 3CXeXor-C1 + 2C2 - 3C3 = X........XX
X”(X) = CXeX + 4CXe0 + 9C3eXorC1 + XXX + 9C3 = X.........iii
XXXXXX i and XX: 3CX - XXX = 0orC2 = 2/3 . CX
Plugging XX in i : C1 + X/X . XX+ CX = X or XX + 5/3 . C3 = X or XX = - 5/X . XX
Plugging C1,C2 in iii : - 5/X . C3 + 4* X/3 . X3 + XXX = X
XXX3 = XXX XX =X/XX
C1 = - X/3 . CX = - 5/3 * 1/XX =-1/X
CX = X/3 . C3 = X/X * 1/XX =1/XX
Hence, the XXXXXXXX for XXX XXXX XXXXX XXXXXXX is:
y(x) = X/XX *e-x - 1/X * eXX + X/10 *e-3x
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