# ma1506

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MA1506 Tutorial 3 Solutions
(1a)
y"+6 y '+9 y = 0

Set y = e λt

λ2 + 6λ + 9 = 0 → λ = −3
→ y = ( A + Bx)e −3 x → y ' = Be −3 x − 3( A + Bx)e −3 x y (0) = 1 ⇒ A = 1 y ' (0) = −1 ⇒ B − 3 A = −1 ⇒ B = 2 → y = (1 + 2 x)e −3 x
(1b)
λ2 − 2λ + (1 + 4π 2 ) = 0 → λ = 1 ± 2πi
→ y = e x [ A cos 2πx + B sin 2πx] y ' = y + e x [−2πA sin 2πx + 2πB cos 2πx] y (0) = −2 ⇒ A = −2 y ' (0) = 2(3π − 1) ⇒ 2(3π − 1) = −2 + 2πB
⇒ B = 3 ⇒ y = e x [−2 cos 2πx + 3 sin 2πx]

(2a)
Try y = Ax 2 + Bx + C y "+2 y '+10 y
= 2 A + 2( 2 Ax + B ) + 10( Ax 2 + Bx + C )
= 25 x 2 + 3
→ 10 A = 25, 4 A + 10B = 0, 2 A + 2B + 10C = 3
→ A = 5 / 2, B = -1, C = 0
5
→ y = x2 − x
2

(2b)
Try y = ( Ax 2 + Bx + C )e 3 x y ' = (2 Ax + B)e 3 x + 3( Ax 2 + Bx + C )e 3 x y" = 2 Ae 3 x + 3(2 Ax + B)e 3 x + 3(2 Ax + B)e 3 x + 9( Ax 2 + Bx + C )e 3 x
9 A − 18 A + 8 A = 1 → A = −1
6 A + 6 A + 9 B − 12 A − 18 B + 8 B = 0 → B = 0
2 A + 3B + 3B + 9C − 6 B − 18C + 8C = 0 → C = −2 y = (− x 2 − 2)e 3 x

(2c) y"− y = 2 x Im e ix

(Im = imaginary part)

if we can solve the complex equation z"-z = 2xe ix then Imz satisfies the above.
Try z = (Ax + B)e ix z' = Ae ix + i(Ax + B)e ix z" = Aie ix + iAe ix − ( Ax + B )e ix z"− z = (2 Ai − Ax − B − Ax − B )e ix = 2 xe ix
→ A = −1
− 2i − 2 B = 0 → B = −i
→ z = (− x − i )e ix = − x cos x + sin x + i[− cos x − x sin x]
Im z = − cos x − x sin x → y = − cos x − x sin x

(2d)
1
1 1
(1 − cos 2 x) = − Re(e 2ix )
2
2 2
1 1
Solve z"+4z = − e 2ix ⇔ Try z = A + Bxe 2ix
2 2
2 ix z" = -4Bxe + 4iBe 2ix → z"+4 z = −4 Bxe 2ix + 4 A + 4 Bxe 2ix + 4iBe 2ix y"+4 y =

→ 4A =

1
1
→ A=
2
8

1
1
= 4iB → B = i
2
8
1 1 2ix 1 z = + ixe = (1 + x(i cos 2 x − sin 2 x))
8 8
8
1 1 y = Re z = − x sin 2 x
8 8

→−

(3a)
Variation of parameters : first solve y"+4y = 0 → y = Acos2x + Bsin2x
Promote A and B to functions A(x), B(x).
Then A(x)cos(2x) + B(x)sin2x is a solution of y"+4y =

1
(1 − cos 2 x)
2

if A(x) and B(x) are

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