EXPLICIT SOLUTIONS TO FIRST ORDER
DIFFERENTIAL EQUATIONS OF SPECIAL TYPES
Abstract. This is a summary of the methods that we have covered so far for solving first order differential equations of the form
y 0 = f (x, y) for certain special classes of functions f (x, y). We
shall always assume that f (x, y) is continuous on some rectangle
a < x < b, c < y < d which may be infinite in either the x direction
or the y direction.
Contents
1. Simplest case: f (x, y) independent of y, y 0 = f (x)
(x)
2. Separation of variables: y 0 = f (x, y) = −M
N (y)
3. Linear equations: y 0 = f (x, y) = −P (x)y + Q(x), or
y 0 + P (x)y = Q(x)
(x,y)
4. Exact equations: y 0 = −M
or M (x, y) + N (x, y)y 0 = 0
N (x,y)
when
∂M (x,y)
∂y
=
∂N (x,y)
∂x
1
2
3
4
1. Simplest case: f (x, y) independent of y, y 0 = f (x)
A differential equation of the form y 0 = f (x) for f (x) a continous
function on a < x < b is solved in one step by integration. The general
solution with free parameter c is
Z x
y(x) =
f (t)dt + c.
(1)
x0
If we add the initial condition y(x0 ) = y0 the solution is
Z x
y(x) =
f (t)dt + y0 .
x0
This follows from the fundamental theorem of calculus, which tells us
that the Riemann integral is an antiderivative, that is
Z x
d
f (t)dt = f (x).
dx x0
1
EXPLICIT SOLUTIONS TO FIRST ORDER DIFFERENTIAL EQUATIONS OF SPECIAL TYPES
2
We can also use the indefinite integral to express the general solution
Z
y(x) = f (x)dx,
which is just another notation for (1).
2. Separation of variables: y 0 = f (x, y) =
−M (x)
N (y)
In this case it is convenient to write the differential equation in the
form
M (x) + N (y)y 0 = 0.
(2)
Again the equation can be solved very quickly using integration. Let
F (x) be an antiderivative of M (x) and G(y), an antiderivative of N (y),
that is
d
d
F (x) = M (x) and
G(y) = N (y).
dx
dy
The solution to (2) is given by the following equation for the function
y(x):
F (x) + G(y(x)) = c.
(3)
That (3) defines a solution to (2) follows from the chain rule for
differentiation,
d
d
dy
G(y(x)) = G(y) .
dx
dy
dx
This implies that any differential function y(x) satisying (3) will satisfy
the differentiated equation
d
dy
d
F (x) + G(y)
M (x) + N (y)y 0 =
dx
dy
dx
d
=
(F (x) + G(y))
dx
dc
because of (3)
=
dx
= 0.
We can write the solution (3) in abbreviated form
Z
Z
M (x)dx + N (y)dy = c.
How do we know there exists a function y(x) satisfying (3). The
Implicit Function Theorem states that if F (x, y) has continous partial
derivatives, (x0 , y0 ) satisfies the equation F (x0 , y0 ) = 0, and ∂F (x∂y0 ,y0 ) 6=
0, then there exists a differentiable function y(x) defined in a neighborhood of x0 satisfying F (x, y(x)) = 0 and y(x0 ) = y0 . In our case
EXPLICIT SOLUTIONS TO FIRST ORDER DIFFERENTIAL EQUATIONS OF SPECIAL TYPES
3
the functions F (x) and G(y) are continously differentiable, so the only
conditions we need to check are F (x0 ) + G(y0 ) = 0 and G0 (y0 ) 6= 0.
Example. Consider the equation
D.E. x + yy 0 = 0 I.C. y(1) = 1.
R
R
The general solution is xdx + ydy = c, or
1 2 1 2
x + y = c.
2
2
Substituting x = 1, y(1) = 1 gives c = 1, multiplying everything by 2
the equation becomes
x2 + y 2 = 2,
(4)
√
This is the equation of a circle of radius
2. Given a point (x0 , y0 )
√
0
(0) = 0,
on the circle other than the points
√ (± 2, 0)√where G (0) = N√
there is a neighborhood of x0 , − 2 < x < 2, where y(x) = 2 − x2
defines a differentiable function of x such that x2 + y(x)2 = 2. The
equation of the circle x2 + y 2 = a2 is a good example for understanding
the implicit function theorem.
3. Linear equations: y 0 = f (x, y) = −P (x)y + Q(x), or
y 0 + P (x)y = Q(x)
The homogeneous linear equation y 0 + P (x)y = 0 can be solved by
separation of variables in the form y1 y 0 + P (x) = 0 and the solution is
Z
R
ln|y| + P (x)dx = c, or |y(x)| = ec e− P (x)dx .
Allowing for y(x) to take negative values we write
y(x) = ke−
R
P (x)dx
.
The solution to the inhomogeneous equation
y 0 + P (x)y = Q(x) can
R
be found by substituting y(x) = v(x)e− P (x)dx . The resulting equation
for v(x) is
v 0 (x)e−
R
P (x)dx
R
= Q(x) or v 0 (x) = Q(x)e
P (x)dx
,
which is solved by integration giving
Z
Z
R
R
R
P (x)dx
− P (x)dx
v(x) = Q(x)e
dx and y(x) = e
Q(x)e P (x)dx dx.
Example:
D.E. y 0 + 2xy = x,
I.C. y(0) = 1.
EXPLICIT SOLUTIONS TO FIRST ORDER DIFFERENTIAL EQUATIONS OF SPECIAL TYPES
4
R
For this example RP (x) = 2x and Q(x) = x. Then P (x)dx =
R
2
2xdx = x2 and e P (x)dx = ex .
Z
Z
R
1 2
2
P (x)dx
v(x) = Q(x)e
dx = xex dx = ex + c.
2
This is where the arbitrary constant in the general solution appears.
The general solution is then
Z
R
R
− P (x)dx
y(x) = e
( Q(x)e P (x)dx dx)
2
2 1
= e−x ( ex + c)
2
1
2
=
+ ce−x .
2
To satisfy the initial condition we substitute y(0) = 1 and get 1 =
2
1
+ ce0 = 12 + c so c = 21 and y(x) = 21 (1 + ex .
2
An equivalent way of solving a first order linear differential equation
isR to multiply the whole equation by the so-called “integrating factor”
e P (x)dx giving
R
e
R
P (x)dx 0
y + P (x)e
P (x)dx
R
y = Q(x)e
P (x)dx
.
(5)
The left hand side of this equation is a derivative
R
(e
P (x)dx
R
y)0 = e
R
P (x)dx 0
y + P (x)e
P (x)dx
y.
So equation (5) becomes
R
(e
P (x)dx
R
y)0 = Q(x)e
P (x)dx
,
and
is solved by integration. In other words, the
R
R equation says that
e P (x)dx y is a function whose derivative is Q(x)e P (x)dx ; hence,
Z
R
R
P (x)dx
e
y = Q(x)e P (x)dx dx,
where the indefinite integral on the right contains an arbitrary constant.
4. Exact equations: y 0 =
when
−M (x,y)
N (x,y)
∂M (x,y)
∂y
or M (x, y) + N (x, y)y 0 = 0
=
∂N (x,y)
∂x
The basic idea here is to find a condition that tells us when the
expression M (x, y) + N (x, y)y 0 looks like a derivative with respect to x
EXPLICIT SOLUTIONS TO FIRST ORDER DIFFERENTIAL EQUATIONS OF SPECIAL TYPES
5
of a function F (x, y(x)), that is
∂F (x, y) ∂F (x, y) 0
+
y
∂x
∂y
d
=
F (x, y(x)).
dx
M (x, y) + N (x, y)y 0 =
This amounts to finding a function F (x, y) such that
∂F (x, y)
= M (x, y) and
∂x
∂F (x, y)
= N (x, y)
∂y
(6)
In addition, we want a proceedure for finding the function F (x, y) when
possible. We saw in class that the necessary and sufficient condition
for the existence of such a function is the “integrability condition”
∂M (x, y)
∂N (x, y)
=
.
∂y
∂x
(7)
Note: Equation (7) is always satified when M (x, y) is independent
of y and N (x, y) is independent of x, which is exactlty the case when
we can separate variables and solve as in Section 2 with a function of
the form F (x) + G(y).
In general, when the integrability condition (7) is satisfied, we can
find the function F (x, y) in the form
Z
F (x, y) = M (x, y)dx + g(y),
R
where M (x, y)dx means an antiderivative of M (x, y) relative to differentiation in x. That is, we first find some function F0 (x, y) such that
∂F0 (x,y)
= M (x, y) and then correct it by adding a function g(y) of y
∂x
(x,y)
= N (x, y). Condition (7) guarantees that such
alone to satisfy ∂F∂y
a function exists.
Example: x + y + (x + y 2 )y 0 = 0.
∂M (x, y)
∂(x + y)
∂(x + y 2 )
∂N (x, y)
=
=1=
=
∂y
∂y
∂x
∂x
Z
F (x, y) =
Z
M (x, y)dx + g(y) =
1
(x + y)dx + g(y) = x2 + xy + g(y).
2
Differentiating in y:
∂( 21 x2 + xy + g(y))
= x + g 0 (y)? =?x + y 2 .
∂y
EXPLICIT SOLUTIONS TO FIRST ORDER DIFFERENTIAL EQUATIONS OF SPECIAL TYPES
6
So we must have g 0 (y) = y 2 or g(y) = 13 y 3 . We introduce the arbitrary
constant at the end in the equation
1
1
F (x, y) = x2 + xy + y 3 = c.
2
3
We don’t expect you to solve this cubic equation explicitly for y(x).
Final example: xy 0 + (x2 + 2y) = 0. In this case
∂M (x, y)
∂(x2 + 2y)
=
=2
∂y
∂y
∂N (x, y)
∂(x)
=
6=
∂x
∂x
What to do? If we divide by x we convert the equation to the linear
equation
y 0 + x2 y = −x. Then we remultiply by the “integrating factor”
R
e P (x)dx = e2ln|x| = |x|2 = x2 to get the equation x2 y 0 + 2xy = −x3 or
x2 y 0 + (x3 + 2xy) = 0, which is equivalent to the original equation at
all points where x 6= 0. This new equation does satisfy the integrability
condition and the solution to the differential equation is x2 y + 41 x4 = c,
that is,
1
c
y = 2 − x2 .
x
4
Checking that we have a solution,
c
1
c
1
xy 0 = x(−2 3 − x) = −2 2 − x2
x
2
x
2
and
xy 0 + 2y = −x2 ,
as desired. The point of this example is to show that sometimes playing
with the equation a little gives an equivalent form which can be solved
explicitly.