Midterm solution (Q.5-Q.8)
5. (a)
x
Z
yn = y0 +
f (t, yn−1 )dt
x0
y0 = 1, f (x, y) = y + xy 2 · · · (2pts)
Z x
1
y1 = 1 +
1 + tdt = 1 + x + x2 · · · (3pts)
2
Z0 x
5 2
1
1 + 2t + t + 2t3 + t4 + t5 dt
y2 = 1 +
2
4
0
5 3 1 4 1 5
1
2
= 1 + x + x + x + x + x + x6 · · · (5pts)
6
2
5
24
(b) Let u = 1/y. Then,
u0 = −y 0 /y 2
and the ODE is
−u0 /u2 = 1/u + x/u2
u0 + u = −x · · · (5pts)
Z
u = e−x
ex (−x)dx + c = e−x (ex − xex + c) = 1 − x + ce−x
y=
1
1 − x + ce−x
From the initial condition,
y=
6.
1
· · · (5pts)
1−x
1 0
1
y + 1 − 2 y = 0.
x
4x
R
1
1
p= ,
e p = e− log x = .
x
x
1
U=
= csc2 x · · · (5pts)
sin2 x
u = − cot x
sin x
cos x
y = uy1 = y1 · √ = − √
x
x
y 00 +
The general solution is
cos x
sin x
y = c1 √ − c2 √ · · · (5pts)
x
x
7. (a) The homogeneous equation is
y 00 + y = 0
The characteristic equation corresponding to it is
λ2 + 1 = 0, ∴ λ = ±i
∴ y = c1 eix + c2 e−ix or y = c1 cos x + c2 sin x · · · (2pts)
W = cos x cos x − sin x(− sin x) = 1 · · · (3pts)
A particular soluton is
Z
Z
yp = − cos x sin tan x + sin x cos x tan x = − cos x log | sec x + tan x| + sin x cos x
∴ y = (c1 − log | sec x + tan x|) cos x + c2 sin x · · · (5pts)
2
(b) The homogeneous equation is
x2 y 00 − 5xy 0 + 5y = 0
The auxiliary equation is
m2 − 6m + 5 = 0
∴ m = 1 or m = 6
y = c1 x + c2 x5 · · · (5pts)
Let y = px log x + q Then
y 0 = p log x + p
y 00 = p/x
So,
px − 5px log x − 5px + 5px log x + 5q = x
1
q=0
∴p=−
4
1
∴ y = c1 x + c2 x5 − x log x · · · (5pts)
4
8. The characteristic equation corresponding to it is
λ3 + 1 = 0 · · · (4pts)
√
√
1−i 3
1+i 3
or λ = ω2 =
∴ λ = −1 or λ = ω1 =
2
2
∴ y = c1 e−x + c2 eω1 x + c3 eω2 x · · · (5pts)
From the initial condition,
c1 + c2 + c3 = 2
−c1 + ω1 c2 + ω2 c3 = −1
· · · (1pts)
c1 − ω2 c2 − ω1 c3 = −1
∴ c1 =
5ω1 + 4
5ω2 + 4
2
, c2 =
, c3 =
3
3(ω2 + 1)
3(ω1 + 1)