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Mathematics (MA-002)
Assignment 2: Differential Equations
Spring Semester 2015-16
1. Using variation of parameter method solve the initial value problem: y 00 + 2y 0 − 3y = tet ,
y(0) = −1/64, y 0 (0) = 59/64.
2. Consider the differential equation
t2 y 00 + 3ty 0 − 3y = 0, t > 0.
(a) Determine r so that y = tr is a solution.
(b) Use (a) to find a fundamental set of solutions.
(c) Use the method of variation of parameters for finding a particular solution to
t2 y 00 + 3ty 0 − 3y =
1
, t > 0.
t3
3. Apply the method of variation of parameters to solve
(i) y 00 + 2y 0 + y = e−t log(t),
(ii) y 00 + y = tan(t),
(iii) y 00 + n2 y = sec(nx),
(iv) y 00 − 2y 0 + y =
et
1+t2
+ 3et
(v) y 00 − 2y 0 + y = tet log(t), t > 0
(vi) y 00 + 4y = 4 cosec2 (t)
(vii) t2 y 00 − (t + 2)ty 0 + (t + 2)y = 2t3
4. Solve the following differential equations by changing the dependent variable (Normal form)
2
(i) y 00 − 2 tan(x)y 0 + 5y = sec(x)ex , (ii) y 00 − 4xy 0 + (4x2 − 1)y = −3ex sin(2x)
(iii) y 00 − x−1/2 y 0 + (1/4)x−2 (x + x1/2 − 8)y = 0
(iv) x2 y 00 − 2(x2 + x)y 0 + (x2 + 2x + 2)y = 0,
(v) (1 − x2 )y 00 − 4xy 0 − (1 + x2 )y = x.
5. Solve the following differential equation by changing independent variable;
x
(i) y 00 + y 0 tan(x) + y cos2 (x) = 0, (ii) y 00 − (1 + 4ex )y 0 + 3e2x y = e2(x+e ) ,
(iii) cos(x)y 00 + sin(x)y 0 − 2 cos3 (x)y = 2 cos5 (x),
(iv) y 00 + (tan(x) − 1)2 y 0 − n(n − 1) sec4 (x)y = 0.
6. Find the series solution of the following equations:
(i) (1 − x2 )y 00 − 2xy 0 + p(p + 1)y = 0, (ii) (1 − x2 )y 0 = 2xy
(iii) y 00 − 3y 0 + 2y = 0,
(iv) y 00 − 4xy 0 + (4x2 − 2)y = 0,
(v) (x2 − 1)y 00 + 3xy 0 + xy = 0, y(2) = 4, y 0 (2) = 6.
(7) Solve the following differential equations in series (Frobenius Method)
(i) 9x(x − 1)y 00 − 12y 0 + 4y = 0, (ii) 2x2 y 00 − xy 0 + (1 − x2 )y = 0
(iii) (2x + x3 )y 00 − y 0 − 6xy = 0
(iv) 4xy 00 + 2y 0 + y = 0
(v) x2 y 00 + xy 0 + (x2 − n2 )y = 0.
Answers
1. y =
et
2
64 (8t
− 4t + 15) − 14 e−3t
(b) {y1 = t, y2 = t−3 }
2. (a) r = 1, −3
2
3. (i) (C1 + C2 t)e−t + ( t2 log(t) −
(c) yp =
−1
16 t
− 41 t−3 log t
3t2 −t
4 )e .
(ii) y = C1 cos(t) + C2 sin(t) − cos(t) log(sec(t) + tan(t)).
(iii) y = C1 cos(nx) + C2 sin(nx) +
1
n2
cos(nx) log(cos(nx)) +
x
n
sin(x).
(iv) y = et (C1 + C2 t + 23 t2 ) − 12 et log(1 + t2 ) + tet tan−1 (t).
(v) y = C1 et + C2 tet + 61 t3 et log(t) −
5 3 t
36 t e
(vi) y = C1 cos 2t + C2 sin 2t − 4 log(sint) cos 2t − 4t sin 2t − 4 cos2 t
(vii) y = c1 t + c2 tet − 2t2 .
√
√
4. (i) y = sec(x)[C1 cos(x 6) + C2 sin(x 6) + 17 ex ]
(iii) y = ex
1/2
[C1 x2 + C2 x−1 ]
2
(ii) y = ex [C1 cos(x) + C2 sin(x) + sin(2x)]
(iv) y = xex (C1 x + C2 )
(v) y = (1 − x2 )−1 (C1 sin(x) + C2 cos(x) + x)
5. (i) y = C1 cos(sin(x)) + C2 sin(sin(x))
√
(iii) y = C1 e
2 sin(x)
+ C2 e
p(p+1) 2
2! x
6. (i) y = C0 [1 −
(ii) y = C0 (1 − x2 )−1
√
− 2 sin(x)
+
x
x
(ii) y = C1 e3e + C2 ee − e2e
+ sin2 (x)
x
(iv) y = C1 e−n tan(x) + C2 e(n−1) tan(x)
p(p−2)(p+1)(p+3) 4
x
4!
. . .] + C1 [x −
(p−1)(p+2) 3
x
3!
+
(p−1)(p−3)(p+2)(p+4) 5
x
5!
. . .]
(iii) y = C0 (1 − x2 − x3 − . . .) + C1 (x + 32 x2 + 76 x3 + . . .)
(iv) y = C0 (1 + x2 + 12 x4 + 16 x6 + . . .) + C1 (x + x3 + 21 x5 + 16 x7 + . . .)
(v) y = 4 + 6(x − 2) −
22
3 (x
7. (i) y = C0 [1 + 31 x +
1.4 2
3.6 x
(ii) y = C0 x[1 +
1 2
2.5 x
+
− 2)2 +
169
27 (x
− 2)3 +
− 2)4 · · ·
y = C1 x7/3 [1 +
+ · · · ] when k = 0,
1
4
2.4.5.9 x
334
81 (x
8
10 x
+
8.11 2
10.13 x
+ · · · ] when k = 73 .
y = C1 x1/2 [1 +
1 2
2.3 x
+
y = C1 x3/2 [1 + 38 x2 −
3
4
128 x
+ · · · ] when k = 3/2.
+ · · · ] when k = 1,
1
4
2.3.4.7 x
+ · · · ] when
k = 1/2.
(iii) y = C0 [1 + 3x2 + 35 x4 − · · · ] when k = 0,
(iv) y = C0 cos x
1/2
(v) y = C0 xn [1 −
y = C1 x
−n
[1 −
when k = 0,
1
2
4(n+1) x
1
2
4(1−n) x
y = C1 sin x
1/2
when k = 1/2.
+
1
4
4.8(n+1)(n+2) x
− · · · ] when k = n,
+
1
4
4.8(1−n)(2−n) x
− · · · ] when k = −n.