EE206
Solutions - Assignment 1
1. Evaluate the following derivatives
*(a)
d
dx
√
ln(4x x + 7)
√
1
ln(4x x + 7) = ln(4) + ln(x) + ln(x + 7)
2
√
d
1
3x + 14
1
=⇒
ln(4x x + 7) = +
=
dx
x 2(x + 7)
2x(x + 7)
(b)
d
dy
y tan(y)ey [4]
d
dy
d
(c)
dz
√
y tan(y)ey = tan(y)ey + y sec2 (y)ey + y tan(y)ey
6 + 6z
[5]
ez
d
dz
(d)
d
dt
√
√
6 + 6z
6 + 6z 1
6
√
=−
+
z
z
e
e
2 ez 6 + 6z
−3(2z + 1)
= z√
e 6 + 6z
− ln(cos(t)) [5]
d
dt
− ln(cos(t)) = −
− sin(t)
cos(t)
= tan(t)
2. Evaluate the following integrals
R
*(a) x cot(x2 + 1)dx
u = x2 + 1
Z
du = 2x dx
Z
1
x cot(x + 1)dx =
cot(u)du
2
Z
1
cos(u)
=
du
2
sin(u)
2
v = sin(u)
dv = cos(u) du
Z
1
1
=
dv
2
v
1
1
= ln |v| + c = ln | sin(u)| + c
2
2
1
2
= ln | sin(x + 1)| + c
2
1
(b)
R
√
cos( x)
√
dx
x
[5]
u=
Z
(c)
R
1
t ln t dt
√
x
du =
1 1
√ dx
2 x
√
Z
cos( x)
√
dx = 2 cos(u)du
x
= 2 sin(u) + c
√
= 2 sin( x) + c
[5]
u = ln(t)
Z
du =
1
dt
t
Z
1
1
dt =
du
t ln t
u
= ln |u| + c
= ln | ln(t)| + c
*(d)
R
eax sin(bx)dx
Z
Z
udv = uv −
u = sin(bx)
du = b cos(bx) dx
1
v = eax
a
dv = eax dx
Z
=⇒
e
ax
1
b
sin(bx)dx = sin(bx)eax −
a
a
dv = eax dx
=⇒
e
ax
Z
eax cos(bx) dx
du = −b sin(bx) dx
1
v = eax
a
u = cos(bx)
Z
vdu
Z
1
b 1
b
ax
ax
ax
sin(bx)dx = sin(bx)e −
cos(bx)e +
e sin(bx) dx
a
a a
a
Z
b2
1+ 2
eax sin(bx)dx =
a
Z
a2 + b2
eax sin(bx)dx =
a2
Z
=⇒
eax sin(bx)dx =
1
b
sin(bx)eax − 2 cos(bx)eax
a
a
1
b
sin(bx)eax − 2 cos(bx)eax
a
a
a
b
sin(bx)eax − 2
cos(bx)eax
a2 + b2
a + b2
a sin(bx) − b cos(bx)
= eax
+c
a2 + b2
2
(e)
R
t sin(t)dt [5]
u=t
du = dt
v = − cos(t)
dv = sin(t)
Z
Z
t sin(t)dt = −t cos(t) +
cos(t)dt
= −t cos(t) + sin(t) + c
(f)
R
ln(y)dy [5]
u = ln(y)
du =
dv = dy
v=y
1
dy
y
Z
Z
ln(y)dy = y ln(y) −
1 dy
= y ln(y) − y + c
3. State whether the following differential equations are linear or nonlinear. Give
the order of each equation.
*(a) (1 − x)y 00 − 4xy 0 + 5y = cos x
linear (in y):
2nd order
*(b) (y 2 − 1)dx + xdy = 0
non linear in y:
1st order
linear in x:
1st order
*(c) t5 y (4) − t3 y 00 + 6y = 0
linear in y:
4th order
(d) (sin θ)y 00 − (cos θ)y 0 = 2 [2]
linear in y:
2nd order
4
d3 y
dy
(e) x dx
+ y = 0[2]
3 −
dx
non linear in y:
3rd order
(f) udv + (v + uv − ueu )du = 0 [2]
linear in v, non linear in u:
1st order [2]
d2 u
dr2
+ du
dr + u = cos(r + u)
non linear:
2nd order [2]
2
(h) ẍ − 1 − ẋ3 ẋ + x = 0
(g)
non linear:
2nd order [2]
3
4. Verify that the indicated functions are solutions to the given differential equations and state whether they are implicit or explicit solutions. Assume an
appropriate interval I of definition.
(a) x2 y 00 + xy 0 + y = 0;
Explicit solution
y = cos(ln(x)) [8]
y = cos(ln(x))
sin(ln(x))
y0 = −
x
sin(ln(x))
cos(ln(x))
y 00 =
−
2
x
x2
Using these in the above equation gives:
x2
cos(ln(x))
sin(ln(x))
sin(ln(x))
− x2
−x
+ cos(ln(x))
2
2
x
x
x
= sin(ln(x)) − cos(ln(x)) − sin(ln(x)) + cos(ln(x)) = 0
(b) 2xydx + (x2 − y)dy = 0; −2x2 y + y 2 = 1 [8]
Implicit solution.
−2x2 y + y 2
dy
dy
−2(2x)y − 2x2
+ (2y)
dx
dx
dy
−4xy − 2(x2 − y)
dx
dy
2xy + (x2 − y)
dx
2xydx + (x2 − y)dy
*(c) xy 0 + xy 2 − y = 0;
Explicit solution
y=
=1
=0
=0
=0
=0
2x
x2 +c
2x
+c
2
4x
y2 = 2
(x + c)2
2(x2 + c) − 4x2
y0 =
(x2 + c)2
−2x2 + 2c
=
(x2 + c)2
y=
x2
Using these in the above equation gives:
−2x3 + 2cx
4x3
2x
+
− 2
2
2
2
2
(x + c)
(x + c)
x +c
=
−2x3 + 2cx + 4x3 − 2x(x2 + c)
(x2 + c)2
=
−2x3 + 2cx + 4x3 − 2x3 − 2cx
=0
(x2 + c)2
4
(d)
dX
dt
= (X − 1)(1 − 2X); ln
Implicit solution
2X−1
X−1
= t [8]
2X − 1
=t
ln
X −1
X −1
(X − 1)(2) − (2X − 1)(1) dX
=1
2X − 1
(X − 1)2
dt
2X − 2 − 2X + 1 dX
=1
(2X − 1)(X − 1) dt
dX
−1
=1
(2X − 1)(X − 1) dt
1
dX
=1
(1 − 2X)(X − 1) dt
dX
= (X − 1)(1 − 2X)
dt
(e) y 00 + y = tan x; y = −(cos x) ln(sec x + tan x) [8]
Explicit solution
y = −(cos x) ln(sec x + tan x)
1
y = (sin x) ln(sec x + tan x) − (cos x)
sec x + tan x
= (sin x) ln(sec x + tan x) − (cos x)(sec x)
0
Using sec x =
1
cos x
= (sin x) ln(sec x + tan x) − 1
00
y = (cos x) ln(sec x + tan x) + (sin x)(sec x)
= −y + tan x
Using these in the above equation gives:
y 00 + y = −y + y + tan x = tan x
5
(sec x tan x + sec2 x)
5. Use the Separation of Variables technique to solve the following first order
differential equations.
dy
(a) (1 − x2 ) dx
+ x(y − a) = 0 [8]
dy
+ x(y − a) = 0
dx
dy
= x(a − y)
(1 − x2 )
dx Z
Z
1
x
dy =
dx
a−y
1 − x2
⇒ du = −2xdx ⇒ − 21 du = xdx
Z
Z
1
1
1
dy = −
du
a−y
2
u
1
(−1) ln(a − y) = − ln(u) + c
2
1
ln(a − y) = ln(u 2 ) − c
1
ln(a − y) = ln (1 − x2 ) 2 − c
!
a−y
= −c
ln
1
(1 − x2 ) 2
a−y
1 = C
(1 − x2 ) 2
(1 − x2 )
Let u = 1 − x2
1
y = a − C(1 − x2 ) 2
dy
= e−y + e−2x−y ;
(b) ex y dx
y(0) = 0 [8]
dy
= ey−x e−y + e−2x−y
dx
y dy
e y
= e−x + e−3x
dx
ey y dy = e−x + e−3x dx
Z
Z
y
e y dy = e−x + e−3x dx
ey−x ex y
u=y
y
dv = e dy
du = dy
v = ey
Z
1
ey dy = −ex − e−3x + c
3
1
yey − ey = −ex − e−3x + c
3
yey −
Imposing initial conditions: y(0) = 0
1
1
+ c =⇒ c =
3
3
1 −3x 1
y
x
(y − 1)e = −e − e
+
3
3
−1 = −1 −
6
(c)
dy
dx
= −2x tan y;
y(0) =
π
2
[8]
dy
= −2x tan y
dx
1
dy = −2xdx
tan y
cos y
dy = −2xdx
sin y
du = cos ydy
1
du = −2xdx
Z u
Z
1
du = −2 xdx
u
x2
ln(u) = −2 + c = −x2 + c
2
2
ln(sin y) = −x + c
Let u = sin y;
sin y = e−x
2 +c
y(x) = sin−1 (e−x
Imposing initial conditions: y(0) =
2 +c
π
2
c
y(0) = sin−1 (e ) =
)
π
2
π
⇒ ec = sin( ) = 1
2
2
y(x) = sin−1 (e−x )
dy
*(d) (1 + x2 ) dx
+ y 2 = 0;
(1 + x2 )
Let x = tan(u);
y(0) =
2
π
dy
+ y2 = 0
dx
1 dy
1
=−
2
y dx
1 + x2
Z
Z
1
1
dy
=
−
dx
2
y
1 + x2
dx = sec2 (u)du
−
1
y
1
y
1
y
1
y
1
y
Z
sec2 (u)
du + c
=−
1 + tan2 (u)
Z
sec2 (u)
=
du − c
sec2 (u)
Z
= 1 du − c
=u−c
= arctan(x) − c
y(x) =
1
arctan(x) − c
Imposing initial conditions: y(0) = π2
1
2
π
y(0) =
=
=⇒ c = −
−c
π
2
1
y(x) =
arctan(x) + π/2
7