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Tutorial Sheet 6 ::: MAL101
Higher order ODEs, System of ODEs, Power Series Method
Find the general solution.
(a) y 000 + 3y 00 + 3y 0 + y = 8ex + x + 3 (b) x3 y 000 + x2 y 00 − 2xy 0 + 2y = x3 ln x (c) xy 000 + 3y 00 = ex
Solve the IVPs.
(a) y 000 + 3y 00 + 3y 0 + y = e−x sin x, y(0) = 2, y 0 (0) = 0, y 00 (0) = −1.
(b) x3 y 000 − 3x2 y 00 + 6xy 0 − 6y = 24x5 , y(1) = 1, y 0 (1) = 3, y 00 (1) = 14.
(c) y iv + 10y 00 + 9y = 40 sinh x, y(0) = 0, y 0 (0) = 6, y 00 (0) = 0, y 000 (0) = −26.
Find the real general solution of the following systems.
(a) y10 = −8y1 − 2y2 , y20 = 2y1 − 4y2 .
(b) y10 = −3y1 − y2 + 2y3 , y20 = −4y2 + 2y3 , y30 = y2 − 5y3 .
(c) y10 = −y1 − 4y2 + 2y3 , y20 = 2y1 + 5y2 − y3 , y30 = 2y1 + 2y2 + 2y3 .
Solve the following IVPs.
(a) y10 = 2y1 + 2y2 , y20 = 5y1 − y2 , y1 (0) = 0, y2 (0) = −7.
(b) y10 = −14y1 + 10y2 , y20 = −5y1 + y2 , y1 (0) = −1, y2 (0) = 1.
Solve the following systems of equations.
(a) y10 = y2 + e3t , y20 = y1 − 3e3t (b) y10 = −2y1 + y2 , y20 = −y1 + et
(c) y10 = 3y1 + y2 − 3 sin 3t, y20 = 7y1 − 3y2 + 9 cos 3t − 16 sin 3t
Solve the following IVPs.
(a) y10 = y2 − 5 sin t, y20 = −4y1 + 17 cos t, y1 (0) = 5, y2 (0) = 2.
(b) y10 = y1 + 4y2 − t2 + 6t, y20 = y1 + y2 − t2 + t − 1, y1 (0) = 2, y2 (0) = −1.
(c) y10 = 5y1 + 4y2 − 5t2 + 6t + 25, y20 = y1 + 2y2 − t2 + 2t + 4, y1 (0) = 0, y2 (0) = 0
Determine the radius of convergence of the following power series.
P
P
P
P∞ (n+3)2 n
n(n+1)
(2n)!
n 4n
n
(b) ∞
(a) ∞
(x − 3)2n (c) ∞
n=1
n=0 (−1) x
n=0 (2n+2)(2n+4) x (d)
n=4 (n−3)4 x
2n
Solve the following using the power series method.
(a) y 00 + xy = 0
(b) y 00 − xy 0 + y = 0
(c) y 00 − y 0 = 0
(d) (2x2 − 3x + 1)y 00 + 2xy 0 − 2y = 0
(a) Prove the so called Rodrigue’s formula for the Legendre polynomial.
1 dn
Pn (x) = n
[(x2 − 1)n ]
(?)
n
2 n! dx
Hint: Expand (x2 − 1)n using the binomial expansion and then differentiate it n times
term by term.
(b) Use (?) to calculate P0 (x), P1 (x), P2 (x), P3 (x) and P4 (x).
(a) Show that
∞
X
1
√
=
Pn (x)tn
(??)
1 − 2xt + t2
n=0
Hint: Use the binomial series for [1 − t(2x − t)]−1/2 and show that the coefficient of tn is
Pn (x).
(b) Using (??) show that Pn (1) = 1, Pn (−1) = (−1)n , P2n+1 (0) = 0, P2n (0) = (−1)n 1·3···(2n−1)
.
2n n!
(c) Differentiate (??) with respect to t to show that
∞
∞
X
X
n
2
(x − t)
Pn (x)t = (1 − 2xt + t )
nPn (x)tn−1
n=0
n=1
Equate the coefficients of tn to obtain the following recursion formula.
(n + 1)Pn+1 (x) = (2n + 1)xPn (x) − nPn−1 (x)
(†)
(d) Assuming that you know P0 (x) = 1 and P1 (x) = x, use (†) to calculate P2 (x), P3 (x), P4 (x)
and P5 (x).
(11) Use the Frobenius method to find a basis of solutions.
(a) xy 00 + 2y 0 − xy = 0
(b) x2 y 00 + 4xy 0 + (x2 + 2)y = 0
(c) xy 00 + (2x + 1)y 0 + (x + 1)y = 0
(d) (x2 + x)y 00 + (4x + 2)y 0 + 2y = 0
::: END :::