Download 2 methods for solving first order odes

CHAPTER
2
METHODS FOR SOLVING FIRST
ORDER ODES
2.1
2.1.1
Separable First Order ODE
Separable Equations and How to Solve Them
Separable First Order ODE
A first differential equation is called separable if it can be rewritten in the
form
g(y) dy = f (x) dx
(2.1)
where g(y) is a function of y (which includes the case of g(y) equalling a
constant) and f (x) is a function of x (which includes the case f (x) equalling
a constant).
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CHAPTER 2. METHODS FOR SOLVING FIRST ORDER ODES
Note that if a differential equation can be rewritten in the form of equation
(2.1.1) by changing the names of the variables, then it is separable. Not all
first order differential equations are separable. The reader should verify that
dy
= x + y is not separable.
dx
Solving a separable first order differential equation amounts to integrating
both sides with respect to their respective variables as shown below.
Solving Separable First Order ODE
Suppose that G(y) is an antiderivative of g(y) and that F (x) is an antiderivative of f (x). Then
G(y) = F (x) + C
solves the separable differential equation
g(y) dy = f (x) dx
Proof: By implicitly differentiating
G(y) = F (x) + C
with respect to x we get
G0 (y)
or
dy
= F 0 (x)
dx
g(y) dy = f (x) dx ¤
Example 2.1 Find all solutions to
dy
= 3y
dx
Solution: We rewrite the DE in differential notation and note that it is
separable (with g(y) = y1 and f (x) = 3).
1
dy = 3 dx
y
(Note that this is only valid, so long as y 6= 0). Next, we antidifferentiate
both sides:
Z
Z
1
dy = 3 dx
y
2.1. SEPARABLE FIRST ORDER ODE
13
and obtain
ln |y| = 3x + C,
dy
= 3y, where C is an arbitrary constant.
dx
This solution can be solved explicitly for y as
which is an implicit solution to
|y| = e3x+C
so
|y| = eC e3x
or
|y| = Ke3x ,
where K = eC is an arbitrary positive constant. Solving for y we obtain
y = ±Ke3x .
Noting that ±K is an arbitrary non-zero constant, we relabel it as K, dropping the condition that K be positive, or
y = Ke3x .
We can see by simply plugging into the DE that the constant function
y = 0 also solves the DE, so
y = Ke3x ,
where K is any arbitrary constant yields a solution to the DE.
¤
Note 1: In the previous example, when writing the DE in differential
notation, we saw that y 6= 0, but in fact we saw that the constant function
y = 0 itself solved the DE. This is the case in general, for if y = c is a zero
1
1
of g(y)
(i.e., g(c)
= 0) then y = c solves g(y) dy = f (x) dx.
Note 2: In the previous example, after antidifferentiating both sides,
only ONE constant of integration is required (why?).
dy
= xy 2 ; y(1) =
Example 2.2 Find a solution to the initial value problem
dx
2
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CHAPTER 2. METHODS FOR SOLVING FIRST ORDER ODES
Solution: We rewrite the DE in differential notation and note that it is
separable.
1
dy = x dx
y2
(Note that this is only valid, so long as y 6= 0). Next, we antidifferentiate
both sides:
Z
Z
1
dy = x dx
y2
and obtain
1
−y −1 = x2 + C
2
Plugging in the initial condition, we obtain
1
1
− = 12 + C
2
2
or
C = −1.
So
1
−y −1 = x2 − 1
2
solves the initial value problem. This can be solved explicitly for y as
y=
1
¤
1 − 21 x2
Note that in the above example, we could have solved explicitly for y
first, then obtained C.
2.1.2
Changing Variables to a Separable Equation
Often in mathematics, a change of variables can be used to transform a
problem into one that can more readily be solved. We provide a few examples
that can be solved by a change of variables.
Example 2.3 Solve the DE
y
y
dy
= sin( ) +
dx
x
x
2.1. SEPARABLE FIRST ORDER ODE
15
Solution: Note that the DE is not separable. Consider the change of variable
v = xy (or vx = y).
Differentiating vx = y with respect to x we see that
v+
dv
dy
x=
dx
dx
So by substitution:
v+
dv
x = sin(v) + v
dx
We obtain
dv
x = sin(v),
dx
which is separable.
csc(v) dv =
1
dx,
x
and
− ln |csc v + cot v| = ln |x| + C
So an implicit solution to the original DE is given by
¯
y ¯¯
y
¯
− ln ¯csc( ) + cot( )¯ = ln |x| + C ¤
x
x
In general, this technique works if
Specifically,
dy
dx
can be expressed as a function of xy .
Suppose there is a differentiable function F (v) so that
dy
= F (y/x)
dx
Then the change of variables v = xy (or vx = y) will transform the differential
equation into a separable differential equation in v and x.
In Example (2.3), the function F (v) = sin v + v.
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CHAPTER 2. METHODS FOR SOLVING FIRST ORDER ODES
Exercises
Solve each of the following:
1.
dz
= zt
dt
2.
dy
x
=
dx
y
3.
dy
= (y 2 + 1)x
dx
4.
dy
= y(y + 1)
dx
5. z 0 = z 2
6.
dz
= t sin z
dt
7. x2 y dx + y 3 x dy = 0
8.
³z ´
dz ³ z ´2
=
+2
dx
x
x
Find solutions for the following initial value problems
9.
dy √
= y; y(1) = 4
dx
10.
dy
= 3yx; y(1) = 2
dx
11.
dy
= 3yx; y(1) = 0
dx
12. θ dr = dθ; r(π) = 1
2
13. y dy − 2x csc y dx = 0; y(1) =
r
π
2