MATH 2280-002
Lecture Notes: 01/11/2013
Math 2280 Section 002 [SPRING 2013]
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Separable Differential Equations
Definition. A first-order differential equation is called separable if it can be written in the form
g(y)
dy
= f (x).
dx
Examples. Which of the following are separable?
• y ′ = 2xy + y? Yes. The DE can be rewritten as g(y)y ′ = f (x), where g(y) =
f (x) = 2x + 1.
• y ′ = exy ? No.
• y ′ = ex+y ? Yes. The DE can be rewritten as e−y y ′ = ex .
• xy ′ + xy = 0? Yes. The DE can be rewritten as y1 y ′ = − x12 .
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y
and
• y ′ = x + y? No. (We can only use multiplication and division to separate the variables.)
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Separation of Variables
Back in calculus, we learned a technique called “separation of variables” for solving separable DE.
Let’s first look at the theory behind this technique.
THEORY:
Let F (x) be an antiderivative of f (x), and let G(y) be an antiderivative of g(y). If
y(x) is a solution to the differential equation
g(y)
dy
= f (x),
dx
then by the chain rule
d
d
G(y(x)) = g(y(x))y ′ (x) = f (x) =
F (x).
dx
dx
Because two antiderivatives only differ by a constant, we conclude that
G(y) = F (x) + C (C is a constant).
dy
= f (x), we just need to integrate
That means that if we have a separable differential equation g(y) dx
g(y) with respect to y and f (x) with respect to x to arrive at an expression that only involves y and
x terms (no derivatives).
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MATH 2280-002
TECHNIQUE: How
Lecture Notes: 01/11/2013
does this theory flesh out in practice?
dy
1. Treat dx
as a quotient of the differentials dy and dx. Multiply and divide (do
not add/subtract) to separate the variables. So
dy
f (x)
=
becomes g(y)dy = f (x)dx.
dx
g(y)
2. Integrate both sides.
Z
Z
g(y)dy = f (x)dx becomes G(y) = F (x) + C,
so we recover the expression we know has to hold because of our theory.
Example. Solve the differential equation
√ dy
1
= 3
.
2 x
dx
y + 4y
1. Separate the variables.
2. Integrate both sides.
1
(y 3 + 4y)dy = √ dx
2 x
Z
3
(y + 4y)dy =
Z
1
√ dx
2 x
√
y4
+ 2y 2 = x + C
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Since y is not solved for in terms of x, our solution is called an implicit solution of the
differential equation. In this case, initial conditions may allow us solve for y explicitly in terms
of x.
Example. Solve the differential equation
dy
= ky.
dx
1. Separate the variables.
2. Integrate both sides.
1
dy = kdx
y
Z
Z
1
dy = kdx
y
ln|y| = kx + C
y = Aekx ,
where A is either ±eC . Since y is not solved for in terms of x, our solution is called an implicit
solution of the differential equation. In this case, initial conditions may allow us solve for y
explicitly in terms of x.
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MATH 2280-002
Lecture Notes: 01/11/2013
Question. This method does not find the solution y(x) = 0! What’s going on?
Looking over our method, we notice that we assume y 6= 0 because we divided by y.
As we noticed in the above example, we need to be careful when we use separation of variables,
because this method may not find every solution.
Warning. Separation of variables may not find every solution to
dy
= g(y)f (x)
dx
because the method assumes g(y) 6= 0.
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