Solutions to Homework 2, Introduction to Differential Equations, 3450:335-003, Dr.
Montero, Spring 2009
Problem 1. The differential equation below is exact. Find a function F(x,y) whose
level curves are solutions to the differential equation
y
dy
−x=0
dx
Solution. Using the notation from class, M(x, y) = −x and N(x, y) = y. Hence
∂N
∂M
(x, y) = 0 =
(x, y),
∂y
∂x
so the equation is indeed exact. Hence we integrate
Z
x2
F (x, y) = M(x, y) dx + g(y) = − + g(y).
2
Next, we impose the condition
y2
∂F
(x, y) = g ′(y) = y so g(y) = .
∂y
2
We conclude that
−x2 + y 2
.
F (x, y) =
2
Problem 2. Use the ”mixed partials” check to see if the differential equation below
is exact. If it is exact find a function F (x, y) whose level curves are solutions to the
differential equation
dy
(−4xy 2 + y) + (−4x2 y + x)
=0
dx
Solution. As before we let
M(x, y) = −4xy 2 + y and N(x, y) = −4x2 y + x.
From here we compute
∂M
∂N
= −8xy + 1 and
= −8xy + 1.
∂y
∂x
The equation is then exact. We integrate for example
Z
Z
F (x, y) = N(x, y) dy + h(x) = (−4x2 y + x) dy + h(x) = −2x2 y 2 + xy + h(x).
Next we use the condition
∂F
∂x
= M(x, y) to obtain
∂F
= −4xy 2 + y + h′ (x) = M(x, y) = −4xy 2 + y.
∂x
This says h′ (x) = 0 so we take h(x) = 0. Our solution is
F (x, y) = −2x2 y 2 + xy.
Problem 3. Use the ”mixed partials” check to see if the differential equation below
is exact. If it is exact find a function F (x, y) whose level curves are solutions to the
differential equation
(4ex sin(y) − 3y) + (−3x + 4ex cos(y))
dy
= 0.
dx
Solution. Again, we set
M(x, y) = 4ex sin(y) − 3y and N(x, y) = −3x + 4ex cos(y),
and compute
∂M
∂N
= 4ex cos(y) − 3 and
= −3 + 4ex cos(y).
∂y
∂x
The equation is exact. Then we integrate for example
Z
Z
F (x, y) = N(x, y) dy+h(x) = (−3x+4ex cos(y)) dy+h(x) = −3xy+4ex sin(y)+h(x).
The next step is to use the condition
∂F
∂x
= M. This means
−3y + 4ex sin(y) + h′ (x) = 4ex sin(y) − 3y.
Again we obtain h′ (x) = 0 so h(x) = 0. Our solution is
F (x, y) = −3xy + 4ex sin(y).
Problem 4. Check that the equation
x2 y 3 + x(1 + y 2)y ′ = 0.
is not exact but becomes exact when multiplied by the integrating factor
µ(x, y) = 1/(xy 3).
Solve the differential equation.
Solution. We let
M(x, y) = x2 y 3 and N(x, y) = x(1 + y 2 ).
We compute
∂M
∂N
= 3x2 y 2 and
= 1 + y 2.
∂y
∂x
The equation is not exact. If we now multiply the equation by µ(x, y) =
equation becomes
1
1 dy
x+
+
= 0.
y 3 y dx
1
,
xy 3
the
We redefine M and N as follows:
M(x, y) = x and N(x, y) =
1
1
+
.
y3 y
This means
∂M
∂N
= 0 and
= 0,
∂y
∂x
so the equation is now exact. We integrate for instance
Z
Z 1
1
1
dy + h(x) = − 2 + ln(y) + h(x).
+
F (x, y) = N(x, y) dy + h(x) =
3
y
y
2y
The condition
∂F
∂x
= M says h′ (x) = x so h(x) =
F (x, y) = −
x2
.
2
This means that our solution is
x2
1
+
ln(y)
+
.
2y 2
2
Problem 5. Find an explicit or implicit solutions to the differential equation
(x2 − 3xy) + x
dy
= 0.
dx
Hint: Try to find an integrating factor that depends only on one variable.
Solution. We let
M(x, y) = x2 − 3xy and N(x, y) = x.
We compute
∂M
∂N
= −3x and
= 1.
∂y
∂x
The equation is not exact. We seek an integrating factor that depends on one variable
only. Let us recall that this is possible if either
∂M
∂y
−
N
∂N
∂x
is independent of y
or
∂N
∂x
−
M
∂M
∂y
is independent of x.
Here
∂M
∂y
−
N
∂N
∂x
=
−3x − 1
x
is independent of y, so we solve
1 dµ
−3x − 1
=
.
µ dx
x
This is a separable equation. We integrate directly to obtain
ln(µ) = −3x − ln(x),
so
e−3x
.
x
We multiply the equation by this integrating factor to obtain the new equation
µ(x) =
e−3x (x − 3y) + e−3x
dy
= 0.
dx
We redefine M, N:
M(x, y) = e−3x (x − 3y) so
∂M
= −3e−3x ,
∂y
and
∂N
= −3e−3x .
∂x
The equation is now exact. We integrate for instance
Z
Z
F (x, y) = M(x, y) dx + g(y) = e−3x (x − 3y) dx + g(y).
N(x, y) = e−3x so
We need the integral
Z
xe−3x dx
which we integrate by parts with u = x and dv = e−3x :
Z
Z
xe−3x 1
xe−3x 1 −3x
−3x
xe
dx = −
e−3x dx = −
+
− e .
3
3
3
9
So far we have
xe−3x 1 −3x
− e
+ ye−3x + g(y).
3
9
= N to obtain
F (x, y) = −
Next we use the condition
∂F
∂y
e−3x + g ′ (y) = e−3x .
We conclude that g ′ (y) = 0 so we take g(y) = 0. The solution we seek is
F (x, y) = −
xe−3x 1 −3x
− e
+ ye−3x .
3
9