Page 27 - Advanced engineering mathematics

1.1 Terminology and Separable Equations 7

Choose k so that
1
y(1) = = 4.
e −1 − k
Solve this equation for k to get
1
−1
k = e − .
4
The solution of the initial value problem is
1
y(x) = .
1
e −x + − e −1
4
It is not always possible to find an explicit solution of a differential equation, in which y is
isolated on one side of an equation and some expression of x occurs on the other side. In such a
case, we must be satisfied with an equation implicitly defining the general solution or the solution
of an initial value problem.

EXAMPLE 1.4
We will solve the initial value problem

(x − 1) 2

y = y ; y(3) =−1.
y + 3
The differential equation itself (not the algebra of separating the variables) requires that y =−3.
In differential form,

y + 3 2
dy = (x − 1) dx
y
or

3
2
1 + dy = (x − 1) dx.
y
Integrate to obtain
1
3
y + 3ln|y|= (x − 1) + k.
3
This equation implicitly defines the general solution. However, we cannot solve for y as an
explicit expression of x.
This does not prevent us from solving the initial value problem. We need y(3) =−1, so put
x = 3 and y =−1 into the implicitly defined general solution to get

3
1
−1 = 2 + k.
3
Then k =−11/3, and the solution of the initial value problem is implicitly defined by
1 3 11
y + 3ln|y|= (x − 1) − .
3 3
Part of this solution is graphed in Figure 1.3.

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October 14, 2010 14:9 THM/NEIL Page-7 27410_01_ch01_p01-42