Lesson 6: Exact equations
Determine whether or not each of equations below are exact. If it is exact find
the solution by two methods.
1. (2x + 4y) + (2x − 2y)y 0 = 0
Solution. The differential equation is not exact since
(2x + 4y)0y = 4,
while
(2x − 2y)0x = 2.
2. (3x2 − 2xy + 2)dx + (6y 2 − x2 + 3)dy = 0
Solution. The differential equation is exact since
(3x2 − 2xy + 2)0y = −2x,
(6y 2 − x2 + 3)0x = −2x.
Let us find a solution.
Method 1. Integrating the first equation, we have
Z
(3x2 − 2xy + 2)dx = x3 − x2 y + 2x + h(y).
To find h(y) we have
−x2 + h0 (y) = 6y 2 − x2 + 3.
Therefore
h0 (y) = 6y 2 + 3,
and
Z
h(y) =
(6y 2 + 3)dy = 2y 3 + 3y.
Therefore, the solution is
x3 − x2 y + 2x + 2y 3 + 3y = c.
Method 2. By integrating on curve (0,0) (x, y) we have
Z x
Z y
2
(3u − 2u × 0 + 2)du +
(6v 2 − x2 + 3)dv
0
0
3
3
2
= x + 2x + 2y − x y + 3y = c.
3. (2xy 2 + 2y) + (2x2 y + 2x)y 0 = 0
Solution.
(2xy 2 + 2y)0y = 4xy + 2,
(2x2 y + 2x)0x = 4xy + 2.
That is the differential equation is exact. Find a solution by two methods.
Method 1. Integrating the first equation we have
Z
(2xy 2 + 2y)dx = x2 y 2 + 2xy + h(y).
1
2
Let us find h(y). We have:
2x2 y + 2x + h0 (y) = 2x2 y + 2x.
That is
h0 (y) = 0,
and the solution is
x2 y 2 + 2xy = c.
Method 2. By integrating on the curve we have
Z x
Z y
2
(2u × 0 + 2 × 0)du +
(2x2 v + 2x)dv = x2 y 2 + 2xy = c.
0
0
4.
dy
ax + by
=−
.
dx
bx + cy
Solution. Rewrite
ax + by
+ y 0 = 0.
bx + cy
Then,
³ ax + by ´0
bx + cy
x
6= 0
in general, while
(1)0y = 0.
That is the differential equation is not exact. With what c it is exact?
5. (ex sin y − 2y sin x)dx + (ex cos y + 2 cos x)dy = 0
Solution.
(ex sin y − 2y sin x)0y = ex cos y − 2 sin x,
(ex cos y + 2 cos x)0x = ex cos y − 2 sin x.
Therefore, the differential equation is exact.
Method 1.
Z
(ex sin y − 2y sin x)dx = ex sin y + 2y cos x + h(y).
Next,
ex cos y + 2 cos x + h0 (y) = ex cos y + 2 cos x.
Therefore,
h0 (y) = 0,
and
ex sin y + 2y cos x = c
is a solution.
Method 2.
Z
x
0
Z
y
(eu × sin 0 − 2 × 0 sin u)du +
(ex cos v + 2 cos x)dv = ex sin y + 2y cos x = c.
0
6. (ex sin y + 3y)dx − (3x − ex sin y)dy = 0.
3
Solution.
(ex sin y + 3y)0y = ex cos y + 3
−(3x − ex sin y)0x = −3 + ex sin y
That is the differential equation is not exact.
7. (yexy cos 2x − 2exy sin 2x + 2x)dx + (xexy cos 2x − 3)dy = 0.
Solution.
(yexy cos 2x − 2exy sin 2x + 2x)0y = exy cos 2x + xyexy cos 2x − 2xexy sin 2x.
(xexy cos 2x − 3)0x = exy cos 2x + xyexy cos 2x − 2xexy sin 2x.
Thus, the differential equation is exact.
Then,
Z
(xexy cos 2x − 3)dy = cos 2xexy − 3y + h(x)
By differentiating in x we have
−2 sin 2xexy + y cos 2xexy + h0 (x) = yexy cos 2x − 2exy sin 2x + 2x
Therefore,
h(x) = x2 ,
and the solution is
cos 2xexy − 3y + x2 = c.
8. (y/x + 6x)dx + (log x − 2)dy = 0, x > 0.
Solution.
(y/x + 6x)0y = 1/x.
(log x − 2)0x = 1/x.
The differential equation is exact.
Therefore (x is positive),
Z
(y/x + 6x) = y log x + 6x2 + h(y).
Differentiation in y yields
log x + h0 (y) = log x − 2,
h(y) = −2y.
The solution is
y log x + 6x2 − 2y = c.
9. (x log y + xy)dx + (y log x + xy)dy = 0, x > 0, y > 0.
Solution.
x
+x
y
y
(y log x + xy)0x = + y
x
The differential equation is not exact.
10.
xdx
ydy
+ 2
= 0.
(x2 + y 2 )3/2
(x + y 2 )3/2
(x log y + xy)0y =
4
Solution.
³
´0
x
3yx
3yx(x2 + y 2 )1/2
=− 2
=−
.
2
2
3/2
(x2 + y 2 )3
y
(x + y )
(x + y 2 )5/2
³
´0
y
3yx
=− 2
.
(x2 + y 2 )3/2 x
(x + y 2 )5/2
Therefore, the differential equation is exact.
Then
Z
Z
d(x2 + y 2 )
xdx
1
=
2
2
3/2
2
(x + y )
(x2 + y 2 )3/2
1
=− 2
+ h(y)
(x + y 2 )1/2
Next,
y
y
+ h0 (y) = 2
.
2
2
3/2
(x + y )
(x + y 2 )3/2
Therefore, the solution is
1
= c.
2
(x + y 2 )1/2
Find an integrating factor and solve the following equations
11. (3x2 y + 2xy + y 3 )dx + (x2 + y 2 )dy = 0.
Solution. We have:
(3x2 y + 2xy + y 3 )0y = 3x2 + 2x + 3y 2 ,
(x2 + y 2 )0x = 2x.
Therefore,
My0 (x, y) − Nx0 (x, y)
= 3.
N (x, y)
Thus, an integrating factor is
³Z
´
I(x, y) = I(x) = exp
3dx = e3x .
Then, the differential equation
e3x (3x2 y + 2xy + y 3 )dx + e3x (x2 + y 2 )dy = 0
is exact.
The solution of the differential equation is
Z y
³
y3 ´
e3x (x2 + v 2 )dv = e3x x2 y +
.
3
0
12.
y 0 = e2x + y − 1.
Solution. Let us write the equation in the form
(e2x + y − 1)dx − dy = 0.
We have
(e2x + y − 1)0y = 1
(−1)0x = 0.
5
Therefore,
My0 (x, y) − Nx0 (x, y)
= −1.
N (x, y)
Then, an integrating factor is
I(x, y) = I(x) = e−x ,
and the differential equation
(ex + e−x y − e−x )dx − e−x dy = 0.
is exact.
13. dx + (x/y − sin y)dy = 0.
Solution. We have
(1)0y = 0,
(x/y − sin y)0x =
1
.
y
Therefore
Nx0 (x, y) − My0 (x, y)
1
= .
M (x, y)
y
Then, an integrating factor is
³ Z dy ´
I(x, y) = I(y) = exp
= y,
y
and the differential equation
ydx + (x − y sin y)dy = 0
is exact.
14. ydx + (2xy − e−2y )dy = 0.
Solution.
(y)0y = 1,
(2xy − e−2y )0x = 2y
Therefore
Nx0 (x, y) − My0 (x, y)
2y − 1
=
.
M (x, y)
y
Then, an integrating factor
Z
2y − 1
I(x, y) =
dy = 2y − log |y|,
y
and the differential equation
(2y 2 − y log |y|)dx + (4xy 2 − 2ye−2y + 2xy log |y| − 2y log |y|e−2y )dy = 0
is exact.
15. ex dx + (ex cot y + 2y sec y)dy = 0.
Solution.
(ex )0y = 0
6
(ex cot y + 2y sec y)0x = ex cot y
Therefore
Nx0 (x, y) − My0 (x, y)
= cot y,
M (x, y)
and the differential equation
cot yex dx + (ex cot2 y + 2y csc y)dy = 0
is exact.
16. [4(x3 /y 2 ) + (3/y)]dx + [3(x/y 2 ) + 4y]dy = 0.
Solution.
[4(x3 /y 2 ) + (3/y)]0y = −8x3 /y 3 − 3/y 2
[3(x/y 2 ) + 4y]0x = 3/y 2
Therefore
Nx0 (x, y) − My0 (x, y)
8x3 /y 3 + 6/y 2
=
= 2y,
M (x, y)
4x3 /y 2 + 3/y
and the differential equation
[8(x3 /y) + 6]dx + [6(x/y) + 8y 2 ]dy = 0
is exact.
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