**Unformatted text preview: **ifferential Equations
Example 2.4: Determine which of the following differential equations is separable.
a) dy
= xy + x
dx b) dy y + 1
=
dx x + 1 c) dy
= y ( y + 1)
dx d) dy
= x+ y
dx Solution:
a) The right side may be factored as x( y + 1) , which meets the condition for separability.
b) The right side is the quotient of a function of y divided by a function of x . Therefore, this
equation is separable.
c) The right side can be thought of as a product y ( y + 1) Ã— 1 , where the constant factor 1 can be
viewed as a function of x . Therefore, the equation is separable.
d) The right side cannot be factored as a product of a function of x and a function of y .
Therefore, this equation is not separable.!
The idea behind the method of separation of variables is to multiply or divide both sides of the
separable differential equation by suitable factors so that all the y terms wind up on the left side,
together with the unknown derivative, and all the x terms wind up on the right side. Following
that step, we then have two big hurdles to overcome. First, we must be able to integrate both sides
of the transformed equation. Second, even after integrating, we will not usually have an explicit
formula for the solution but rather an implicit one. Some tricky algebra may be needed to obtain
an explicit solution. Often, an explicit solution cannot be found. We illustrate with some
examples, beginning with a reprise of Example 2.2.
Example 2.5: Use the method of separation of variables to find the general solution of dy
= y.
dx Solution: ! Although we will eventually work with the differential notation, to understand the method it is
better to write the equation as yâ€² = y . Following the instructions in the previous paragraph, we
divide both side of the equation by y . (It is worth emphasizing that division and/or
multiplication must be used to move factors from one side to the other, never addition or
subtraction.) We then obtain
yâ€²
= 1.
y
19 2 Differential Equations
Now we integrate both sides of the latter equation with respect to x , the independent variable.
This gives
yâ...

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