Description

Patrick Gardocki
3.6: 28
x2y′′ − 3xy′ + 4y = 0; y(1) = 5; y′1) = 2 y′(x) = mxm−1; y′′(x) = m(m − 1)xm−2
(m(m − 1) − 3m + 4)xm = 0
(m2 − 4m + 4)xm = 0
(m − 2)2 = 0 ∴ m1 = m2 = 2
yc(x) = c1x2 + c2x2lnx
y′(x) = 2c1x + 2c2xlnx + c2x y(1) = 5 = c1 y′(1) = 2 = 2c1 + c2
c1 = 5 c2 = −8
y(x) = 5×2 − 8x2lnx
Plot of y(x) = 5x^2 − 8x^2 * ln(x)

3.8: 41
ddt22x −5sin2t + 3cos2t; x(0) = −; x′(0) = 1 + 4x = m2 + 4 = 0 → m1 = −2i, m2 = 2i
xc(t) = c1cos2t + c2sin2t
xp(t) = Atcos2t + Btsin2t x′p(t) = (A + 2Bt)cos2t + (−2A + Bt)sin2t x′′p(t) = (−4At + 4B)cos2t + (−4A − 4Bt)sin2t
4Bcos2t − 4Asin2t = 3cos2t − 5sin2t
4B = 3 → B = − 4A = −5 → A =
xp(t) = tcos2t + tsin2t x(t) = c1cos2t + c2sin2t + tcos2t + tsin2t Given: x(0) = −; x′(0) = 1 x(t) = −cos2t − sin2t + tcos2t + tsin2t
3.9: 12
y′′ + λy = 0; y′(0) = 0; y(π/4) = 0
For λ = 0, y(x) = c1x + c2, y′(0) = y(π/4) = 0 → c1 = c2 = 0
y(x) = 0 For λ = 0, no nontrivial solutions, and is not eigenvalue.
For λ < 0, λ = −α2; y′′ − α2y = 0; m2 − α2 = 0 → m1 = −α, m2 = α y(x) = c1coshαx + c2sinhα → c1 = c2 = 0 → y(x) = 0
For λ < 0, no nontrivial solutions, and is not eigenvalue.
For λ > 0, λ = α2; y′′ + α2y = 0; m2 + α2 = 0 → m1 = −iα, m2 = iα c2 = 0; c1cos = 0
For c1 ̸= 0, nontrivial solutions. cos = 0 → α = 2 + 4n λn = (2 + 4n)2 yn(x) = sin(2 + 4n)x
5.1: 8
1−x = 3−2−x = −1 + 2+3x = −1 + (32)(1+1×2 ) =− 34x + 3×82 − 316×3 + …
2+x 2−x
|x2| < 1 → Interval of Convergence (−2,2)
5.1: 19
5.2: 6
5.2: 19
10.2: 22
10.2: 38
10.4: 14 by using variation of parameters using the matrix exponential formula