Math240
Elementary Differential Equations
Fall 2003
Kansas State University
Exact Equations (Section 1.4):
Solutions for Selected Problems
dy
= 0.
(1) 2xy + x2 dx

Step 1: ∂ [2xy] = 2x; 

∂y
the equation is exact.
∂ £ 2¤

x = 2x 
∂x
Step 2: ∂F = 2xy;
∂x
∂F
= x2 .
∂y
Z
Step 3: F (x, y) =
Step 4: F (x, y) =
Z
2xy dx + C(y) = x2 y + C(y).
e
e
x2 dy + C(x)
= x2 y + C(x).
e
e
Step 5: F (x, y) = x2 y + C(y) = x2 y + C(x)
⇒ C(y) − C(x)
=0
2
e
⇒ C(y) = 0 and C(x) = 0.Step 6: F (x, y) = x y.
x2 y = K or y =
K
x2
µ
¶
1 dy
2
= 0.
(3) e + 2xy + 2x y −
y dx
£
¤
Step 1: ∂ ex + 2xy 2 = 4xy;
∂y ·
¸
∂
1
2
2x y −
= 4xy
∂x
y
x
2







the equation is exact.
Step 2: ∂F = ex + 2xy 2 ;
∂x
∂F
1
= 2x2 y − .
∂y
Z y
¡ x
¢
Step 3: F (x, y) =
e + 2xy 2 dx + C(y) = ex + x2 y 2 + C(y).
¶
Z µ
1
2
e
e
Step 4: F (x, y) =
dy + C(x)
= x2 y 2 − ln |y| + C(x).
2x y −
y
e
Step 5: F (x, y) = ex + x2 y 2 + C(y) = x2 y 2 − ln |y| + C(x)
e
⇒ C(y) − C(x)
= − ln |y| − ex
e
⇒ C(y) = − ln |y| and C(x)
= ex .
2 2
x
Step 6: F (x, y) = x y + e − ln |y|.
x2 y 2 + ex − ln |y| = K
1
(5)
yex + y
dy
=− x
.
dx
e +x
First, we rewrite the equation as
dy
yex + y + (ex + x)
= 0.
dx

∂
x
x
Step 1:

[ye + y] = e + 1; 
∂y
the equation is exact.
∂ x

x

[e + x] = e + 1
∂x
Step 2: ∂F = yex + y;
∂x
∂F
= ex + x.
∂y
Z
Step 3: F (x, y) =
Step 4: F (x, y) =
Z
(yex + y) dx + C(y) = yex + yx + C(y).
e
e
(ex + x) dy + C(x)
= yex + yx + C(x).
e
e
Step 5: F (x, y) = yex + xy + C(y) = yex + xy + C(x)
⇒ C(y) − C(x)
=0
e
⇒ C(y) = 0 and C(x) = 0.
Step 6: F (x, y) = yex + xy.
yex + xy = K or y =
(7) x2 y + 2xy 2
ex
K
.
+x
dy
= 0.
dx
£ ¤
Step 1: ∂ x2 y = x2 ;
∂y
¤
∂ £
2xy 2 = 2y 2
∂x





the equation is not exact.
By dividing the original equation by y, we can make this into a separable equation.
We rewrite the equation as
µ
¶
1
1
dy
2
2 dy
x y + 2xy
= · 0 ⇒ x2 + 2xy
= 0.
y
dx
y
|
{zdx }
Modified Equation
dy
Step 1: x2 + 2xy
= 0 ⇒ 2y dy = −x dx.
dx
1
Step 2: y 2 = − x2 + C.
2
r
1
Step 3: y = ± C − x2
|
{z 2 }
General Solution
2
Step 4: Check for singular solutions: to rewrite the equation, we divided by y
dy
= 0?
Is y(x) = 0 a solution to the original equation x2 y + 2xy 2
dx
Yes, it is.
(notice that it is not a solution of the modified equation).
Is the solution
y(x) = 0 represented in the general solution?
r
1
1
y = ± C − x2 = 0 ⇒ C = x2
2
2
(no good, since C must be a constant.)
No, it is not represented.
y(x)=0 is a singular solution.
r
r
1 2
1
y = C − x , y = − C − x2 , and y = 0.
2
2
(9)
dy
x+1
=
.
dx
y+1
First, we rewrite this equation as
dy
(x + 1) − (y + 1) .
dx

∂

Step 1:

[x + 1] = 0;
∂y
the equation is exact.
∂

[−(y + 1)] = 0 
∂x
Step 2: ∂F = x + 1;
∂x
∂F
= y + 1.
∂y
Z
1
(x + 1) dx + C(y) = x2 + x + C(y).
2
Z
1
e
e
Step 4: F (x, y) = (y + 1) dy + C(x)
= y 2 + y + C(x).
2
1
1
e
Step 5: F (x, y) = x2 + x + C(y) = y 2 + y + C(x)
2
2¢ ¡
¡1 2
¢
e
⇒ C(y) − C(x)
= 2 y + y − 12 x2 + x
1
1
e
⇒ C(y) = y 2 + y and C(x)
= x2 + x.
2
2
¢
1¡ 2
2
Step 6: F (x, y) =
x + y + x + y.
2
¢
1¡ 2
x + y 2 + x + y = K.
2
Step 3: F (x, y) =
3
¡
¢ dy
= 0; y(0) = 1.
(11) 3x2 tan(y) + 1 + x3 sec2 (y) − 1
dx
First: Find all solutions, if the equation is exact.

£
¤
Step 1: ∂ 3x2 tan(y) + 1 = 3x2 sec2 (y); 

∂y
the equation is exact.
¤
∂ £ 3 2

x sec (y) − 1 = 3x2 sec2 (y) 
∂x
Step 2: ∂F = 3x2 tan(y) + 1;
∂x
∂F
= x3 sec2 (y) − 1.
∂y
Z
¢
¡ 2
Step 3: F (x, y) =
3x tan(y) + 1 dx + C(y) = x3 tan(y) + x + C(y).
Z
¡ 3 2
¢
e
e
Step 4: F (x, y) =
x sec (y) − 1 dy + C(x)
= x3 tan(y) − y + C(x).
e
Step 5: F (x, y) = x3 tan(y) + x + C(y) = x3 tan(y) − y + C(x)
e
e
⇒ C(y) − C(x)
= −y − x ⇒ C(y) = −y and C(x)
= x.
3
Step 6: F (x, y) = x tan(y) + x − y.
The solutions satisfy x3 tan(y) + x − y = K.
Second: Use the initial condition.
y(0) = 1 ⇒ 03 · tan(1) + 0 − 1 = K ⇒ K = −1.
x3 tan(y) + x − y = −1.
dy
(13) 2x + y + 2y
= 0; y(0) = 1.
dx
First: Find all solutions, if the equation is exact.

Step 1: ∂ [2x + y] = 1; 

∂y
the equation is not exact.
∂


[2y] = 0
∂x
π
dy
= 0; y(0) = .
(15) cos(y) − x sin(y)
dx
2
First: Find all solutions, if the equation is exact.

∂

Step 1:

[cos(y)] = − sin(y);
∂y
the equation is exact.
∂


[−x sin(y)] = − sin(y)
∂x
Step 2: ∂F = cos(y);
∂x
∂F
= −x sin(y).
∂y
Step 3: F (x, y) = x cos(y) + C(y).
e
Step 4: F (x, y) = x cos(y) + C(x).
e
Step 5: C(y) = 0 and C(x)
= 0.
Step 6: F (x, y) = x cos(y).
4
The solutions satisfy x cos(y) = K.
Second: Use the initial condition.
π
y(0) = ⇒ 0 · cos(π/2) = K ⇒ K = 0.
2
π
So, x cos(y) = 0, which implies that y = (2k + 1) , where k is an integer. Among
2
π
these solutions only y =
satisfies the initial condition.
2
(17)
x+y
dy
=
; y(0) = 1.
dx
2y − x
First: Find all solutions, if we can rewrite the equation as an exact equation.
dy
(x + y) + (x − 2y)
= 0.
 dx
Step 1: ∂ [x + y] = 1; 

∂y
the equation is exact.
∂


[x − 2y] = 1
∂x
Step 2: ∂F = x + y;
∂x
∂F
= x − 2y.
∂y
1
Step 3: F (x, y) = x2 + xy + C(y).
2
e
Step 4: F (x, y) = −y 2 + xy + C(x).
1
e
Step 5: C(y) = −y 2 and C(x)
= x2 .
2
1 2
2
Step 6: F (x, y) = x + xy − y .
2
1 2
The solutions satisfy x + xy − y 2 = K.
2
Second: Use the initial condition.
1
y(0) = 1 ⇒ · 02 + 0 · 1 − 12 = K ⇒ K = −1.
2
1
Hence y 2 − xy − x2 − 1 = 0, which implies that
2
q
¢
¡
x ± x2 − 4 − 12 x2 − 1
1
1√ 2
y=
⇒y = x±
3x + 4.
2
2
2
The initial condition now tells us that we should take the positive root.
1
1√ 2
y = x+
x + 4.
2
2
dy
ex − y
(19)
=
; y(0) = 0.
dx
x + cos(y)
First: Find all solutions, if we can rewrite the equation as an exact equation.
(ex − y) − (x + cos(y))
5
dy
= 0.
dx
Step 1: ∂ [ex − y] = −1;
∂y
∂
[−x − cos(y)] = −1
∂x





the equation is exact.
Step 2: ∂F = ex − y;
∂x
∂F
= −x − cos(y).
∂y
Step 3: F (x, y) = ex − xy + C(y).
e
Step 4: F (x, y) = −xy − sin(y) + C(x).
e
Step 5: C(y) = − sin(y) and C(x)
= ex .
x
Step 6: F (x, y) = e − xy − sin(y).
ex − xy − sin(y) = 1.
6