Description

Patrick Gardocki
1.1.28
Set f(x) = R0x et2dt; f′(x) = dxd R0x et2dt = ex2
dx dy = dxd [e−x2] + dxd [e−x2f(x)] = −2xe−x2 + dx[e−x2]f(x) + e−x2f′(x)
= −2xe−x2 − 2xe−x2 R0x et2dt + e−x2 · ex2
= 1 − 2xe−x2 − 2xe−x2 R0x et2dt
For dx dy + 2xy = 1;
1 − 2xe−x2 − 2xe−x2 R x et2dt + 2xe−x2 + 2xe−x2 R x et2dt = 1
0
∴ 1 = 1
This is a solution.
1.2.22
(1 + y3)y′ = x2; f(x.y) = dx dy = 1+x2y3
df = −3x2y21 + y 332
dy 0

dy df dx and dy are both continuous when y ̸= −1.
For (x0,y0) such that y ̸= −1, there is a unique solution.
2.2.25
Given x2dxdy = y − xy, y(−1) = −1 x2dxdy = y(1 − x) → R dyy = R 1x−2xdx → ln|y| = −x1 − ln|x| + C
If y(−1) = −1,→ y = −1, x = −1
Plug into previous equation to solve for C.
C = −1
ln|y| = −x1 − ln|x| − 1
Solve for y
eln|y| = e−x1−ln|x|−1 → y = 1e−x1−1
x
2.3.13
Given: x2y′ + x(x + 2)y = ex → x2y′ + (x2 + 2x)y = ex
dy 2 ex
+ (1 + )y = EQ1
dx x x2
P xex2
P(x) is continuous on (−∞,0) and f(x) is continuous on (0,∞).
I(x) = eR P(x)dx = eR exe2ln|x| = x2ex; I(x) is continuous on (0,∞)
EQ1 ∗ I(x) = x2exdx dy + (x2ex + 2xe2)y = e2x d 2exy = e2x → x2exy = R e2xdx → x2exy = 1e2x + C
x
dx 2
Solve for y: General Solution: y = 2 x12ex e2x + x2Cex on interval (0,∞)
2.5.18
Given: xdx dy − (1 + x)y = xy2
Substitution: u = y−1; dxdy = −u−2dudx → dudx + 1+xxu = −1
R 1+x
Integrating Factor: e x = elnx+x = xex xexdudx + (ex + xex)u = −xex; dxd xexu = −xex → xexu = R −xexdx = −xex + ex + c → u = −1 + x1 + xce−x
u = y−1 → y = − 1+x11 xc −x + e
2.8.2
Using Equation: P(t) = bP0+(aaP−bP0 0)e−at
Given: N(0) = N0 = 500; N(1) = 1000; lim∞ = ab = 50000
N(t) = 500a
N −at −at −at
Sub in N
N t
3.1.26
For f1(x) = ex/2; f2(x) = xex/2
Linearly Independent if W(f1,f2) ̸= 0
f1 f2
W(f1,f2) = f1′ f2′ = f1 · f2′ − f1′ · f2

1eex/x/22 ex/2xe+x/xe2 x/2 = ex ̸= 0
2
∴ f1andf2 form a set of solutions on interval (−∞,∞).
General Solution: y = c1ex/2 + c2xex/2
3.3.35
Given: y′′′ + 12y′′ + 36y′ = 0; y(0) = 0; y′(0) = 1; y′′(0) = −7
m3+12m2+36m = 0 = m(m2+12m+36) → m1 = 0; m2+12m+36 = 0 = (m+6)2 = 0 → m2 = m3 = −6
y = c1em1x + c2em2x + c3xem3x = c1 + c2e−6x + c3xe−6x y′ = −6c2e−6x + c3(e−6x − 6xe−6x) y′′ = 36c2e−6x + c3(−12e−6x + 36xe−6x)
Sub in y(0) = 0: c1 + c2 = 0
Sub in y′(0) = 1: −6c2 + c3 = 1 → −72c2 + 12c3 = 12
Sub in y′′(0) = −7: 36c2 − 12c3 = −7
Add y′ and y′′: −36c2 = 5 → c2 = −365,c3 = 16,c1 =
y = − e−6x + xe−6x
3.4.15
Given: y′′ + y = 2xsinx
√
For: m2 + 1 = 0; m2 = −1 = ± −1 = ±i → m1 = i; m2 = −i α = 0; β = 1 → yc = c1cosx + c2sinx
Assume: yp = (Ax2 + B)cosx + (Cx2 + D)sinx yp′ = (2Ax + B)cosx − (Ax2 + bx)sinx + (2Cx + D)sinx +
(Cx2 + Dx)cosx
yp′′ = 2Acosx − (2Ax + B)sinx − (2Ax + B)sinx − (Ax2 + bx)cosx + 2Csinx + (2Cx + D)cosx + (2Cx +
D)cosx − (Cx2 + Dx)sinx Subbing into DE:
−4Axsinx = 2xsinx → A =
4Cxcosx = 0 → C = 0
(2A + 2D)cosx = 0 → D = (2B + 2C)sinx = 0 → B = 0
yp = −21x2cosx + xsinx y = yp + yc = c1cosx + c2sinx − 12x2cosx + xsinx
3.4.31
Given: y′′ + 4y′ + 5y = 35e−4x, y(0) = −3, y′(0) = 1
m2 + 4m + 5 = 0 → m1 = −2 + i, m2 = −2 − i ∴ α = −2, β = 1
yc = e−2x(c1cosx + c2sinx)
Assume: yp = Ae−4x; yp′ = −4Ae−4x, yp′′ = 16Ae−4x Subbing into DE:
16Ae−4x − 16Ae−4x + 5Ae−4x = 35e−4x → A = 7
∴ yp = 7e−4x, y = yp + yc y = e−2x(c1cosx + c2sinx) + 7e−4x; y′ = −2e−2x(c1cosx + c2sinx) + e−2x(−c1sinx + c2cosx) − 28e−4x
For y(0) = 0 → c1 = −10 For y′(0) = 1 → c2 = 9 y = e−2x(−10cosx + 9sinx) + 7e−4x