MTHE 225
HOMEWORK SET 7
1. Solve the differential equation: [8 sin y − 9y sin x] dx + [9 cos x + 8x cos y − 10y] dy = 0.
5y
2. Solve the differential equation: 8(1 + ln x) +
dx + [5 ln x − 9y 8 ] dy = 0.
x
7
7
3. Solve the differential equation: y 0 − y = − e−7x y 9 .
8
8
4. Solve the differential equation: y 0 +
3
12
y = 96x4 y 4 .
x
5. Solve the differential equation: (6x + 8y) dx − (8x + 5y) dy = 0.
1
SOLUTION
1.
This differential equation ”looks like” either in homogeneous form or in exact form. But
a quick but smart observation would tell you it is not in homogeneous form (see the following
argument).
(If it is in homogeneous form then: M (x, y) = 9 cos x + 8x cos y − 10y and N (x, y) =
8 sin y − 9y sin x.
Now we notice:
M (tx, ty) = 9 cos (tx) + 8x cos (ty) − 10ty (clearly we can not factor out t)
N (tx, ty) = 8 sin (ty) − 9y sin (tx) (clearly we can not factor out t)
Hence the given differential equation is in not homogeneous form.)
Now let us try to check whether it is in exact form.
If it is in exact form them M (x, y) = 8 sin y − 9y sin x and N (x, y) = 9 cos x + 8x cos y − 10y.
Now Notice:
∂M
∂N
∂M
∂N
= (8 cos y − 9 sin x) and
= (−9 sin x + 8 cos y). Therefore
=
and that
∂y
∂y
∂y
∂y
means the given differential equation is in exact form. That means there exists a function
df
df
f (x, y) such that fx =
= M and fy =
= N.
dx
dy
So,
fx = M = 8 sin y − 9y sin x
Z
Z
⇒ f = fx dx = (8 sin y − 9y sin x) dx = 8x sin y + 9y cos x + g(y)
⇒ fy = 8x cos y + 9 cos x + g 0 (y). But fy = N.
⇒ 9 cos x + 8x cos y − 10y = 8x cos y + 9 cos x + g 0 (y)
⇒ g 0 (y) = −10y
Z
Z
0
⇒ g (y) dy = (−10y) dy
⇒ g(y) = −5y 2 + C
⇒ f = 8x sin y + 9y cos x − 5y 2 + C
⇒ y = 8x sin y + 9y cos x − 5y 2 + C
(implicit solution)
2
2.
This differential equation ”looks like” either in homogeneous form or in exact form. But
a quick but smart observation would tell you it is not in homogeneous form (see the following
argument).
5y
8
.
(If it is in homogeneous form then: M (x, y) = [5 ln x−9y ] and N (x, y) = 8(1 + ln x) +
x
Now we notice:
M (tx, ty) = [5 ln (tx) − 9(ty)8 ] = 5 ln t + 5 ln x − 9t8 y 8 (clearly we can not factor out t)
5ty
5y
N (tx, ty) = 8(1 + ln (tx)) +
= 8(1 + ln t + ln (x)) +
(clearly we can not factx
x
tor out t). Hence the given differential equation is in not homogeneous form.)
Now let us try to check whether it is in exact form.
5y
and N (x, y) = [5 ln x − 9y 8 ].
If it is in exact form them M (x, y) = 8(1 + ln x) +
x
Now Notice:
∂M
5
∂N
5
∂M
∂N
=
and
= . Therefore
=
and that means the given differential
∂y
x
∂y
x
∂y
∂y
df
equation is in exact form. That means there exists a function f (x, y) such that fx =
=M
dx
df
= N.
and fy =
dy
So,
5y
5y
fx = M = 8(1 + ln x) +
= 8 + 8 ln x +
x
x
Z
Z 5y
⇒ fx dx =
8 + 8 ln x +
dx = 8x + 8(x ln x − x) + 5y ln x + g(y) (see remark)
x
⇒ f = 8x + 8x ln x − 8x + 5y ln x + g(y) = 8x ln x + 5y ln x + g(y)
⇒ fy = 5 ln x + g 0 (y) But fy = N.
⇒ 5 ln x − 9y 8 = 5 ln x + g 0 (y)
Z
Z
0
8
0
⇒ g (y) = −9y ⇒ g (y) dy = −9 y 8 dy ⇒ g(y) = −y 9 + C
⇒ f = 8x ln x + 5y ln x − y 9 + C
⇒ y = 8x ln x + 5y ln x − y 9 + C
(implicit solution)
3
Z
Remark: To find
ln x dx use “integration by parts” with u = ln x and dv = dx. That
Z
Z
R
R
1
ln x dx = uv − v du =
means du = dx and v = dv = dx = x. Therefore,
x
Z
Z
1
x ln x − x · dx = x ln x −
dx = x ln x − x.
x
3.
This differential equation clearly a Bernoulli’s differential equation with n = 9. Therefore
du
dy
1
set u = y 1−n = y −8 or y = u−1/8 and hence
= − u−9/8
. Now substitute all these in
dx
8
dx
the given differential equation:
7
7
y 0 − y = − e−7x y 9
8
8
1
7
du 7 −1/8
⇒ − u−9/8
− u
= − e−7x u−9/8 (now multiply both sides by −8u9/8 )
8
dx 8
8
du
⇒
+ |{z}
7 u = 7e−7x
(?)
dx
p(x)
This is a linear differential equation
with p(x) = 7 and hence integrating factor method is
R
7 dx
applicable. Therefore, I.F. = e
= e7x . Now multiply both sides of (?) by this I.F. and
we get:
e7x
du
+ 7e7x = 7
dx
d
⇒
(ue7x ) = 7 ⇒ d(ue7x ) = 7 dx ⇒
dx
Z
7x
d(ue ) = 7
Z
dx
⇒ ue7x = 7x + C
⇒ y −8 e7x = 7x + C
e7x
7x + C
7x 1/8
e
⇒y=
7x + C
(implicit solution)
⇒ y8 =
(explicit solution)
4.
This differential equation clearly a Bernoulli’s differential equation with n = 3/4. Theredy
du
fore set u = y 1−n = y 1/4 or y = u4 and hence
= 4u3
. Now substitute all these in the
dx
dx
given differential equation:
y0 +
3
12
y = 96x4 y 4
x
4
du 12 4
1
+ u = 96x4 u3 (now multiply both sides by u−3 )
dx
x
4
du
3
⇒
+
u = 24x4
(? ?)
dx |{z}
x
⇒ 4u3
p(x)
This is a linear differential equation
with p(x) = x3 and hence integrating factor method is
R 3
applicable. Therefore, I.F. = e x dx = e3 ln x = x3 . Now multiply both sides of (? ?) by this
I.F. and we get:
du
+ 3x2 u = 24x7
dx
Z
Z
d
3
7
3
7
3
⇒
(ux ) = 24x ⇒ d(ux ) = 24x dx ⇒
d(ux ) = 24 x7 dx ⇒ ux3 = 3x8 + C
dx
x3
⇒ y 1/4 x3 = 3x8 + C
(implicit solution)
⇒ y 1/4 = 3x5 + Cx−3
⇒ y = (3x5 + Cx−3 )4
(implicit solution)
5.
WAY 1:
This differential equation ”looks like” either in homogeneous form or in exact form. So let
us first check whether it is in homogeneous form; and if it is not, then we can try to check
whether it is in exact form.
So if it is in homogeneous form then M (x, y) = (8x + 5y) and N = (6x + 8y).
Now notice:
M (tx, ty) = (8tx + 5ty) = t((8x + 5y)) = tM (x, y)
N (tx, ty) = (6tx + 8ty) = t(6x + 8y) = tM (x, y)
So in both cases we could factor out t or t1 . Hence the given differential equation is in homogeneous form. So set y = ux. That means dy = x du + u dx. Now plugging these in the
given differential equation you can proceed like Problem#3 and Problem#4 of homework
Set 6. (But this process could be long!)
WAY 2:
Let us try to check whether it is in exact form. If it is in exact form them M (x, y) = (6x+8y)
and N (x, y) = (8x + 5y).
5
Now Notice:
∂M
∂N
∂M
∂N
= 8 and
= 8. Therefore
=
and that means the given differential equa∂y
∂y
∂y
∂y
df
=M
tion is in exact form. That means there exists a function f (x, y) such that fx =
dx
df
and fy =
= N . Now you can proceed like Problem#1 and Problem#2 of this homework
dy
Set 7.
(That means it is possible to solve a differential equation in several different
ways!)
6