Page 93 - Advanced engineering mathematics

2.5 Euler’s Differential Equation 73

Substitute these into Euler’s equation to obtain the transformed differential equation

Y (t) − Y (t) + AY (t) + BY(t) = 0
or
Y (t) + (A − 1)Y (t) + BY(t) = 0. (2.16)

This is a constant coefficient equation which we know how to solve.
We need not go through this derivation whenever we encounter an Euler equation. The coef-
ficients of equation (2.16) can be read directly from those of the Euler equation. Solve this
transformed equation for Y(t), then replace t = ln(x) to obtain the solution y(x) of the Euler
equation. In doing this, it is useful to recall that, for any number r and for x > 0,
r
x = e r ln(x) .
Furthermore,
e ln(k) = k
for any positive quantity k. Thus, for example,
3
3
e 3ln(x) = e ln(x ) = x .

EXAMPLE 2.21
We will find the general solution of
2

x y + 2xy − 6y = 0.
With A = 2 and B =−6, this Euler equation transforms to
Y (t) + Y (t) − 6Y(t) = 0.

This constant coefficient linear homogeneous equation has general solution
2t
Y(t) = c 1 e −3t + c 2 e .
Replace t = ln(x) to obtain the general solution of the Euler equation:
y(x) = c 1 e −3ln(x) + c 2 e 2ln(x) = c 1 x −3 + c 2 x 2
for x > 0.

EXAMPLE 2.22
2
Consider the Euler equation x y − 5xy + 9y = 0. The transformed equation is

y − 6y + 9y = 0,

3t
3t
with the general solution Y(t) = c 1 e + c 2 te . The Euler equation has the general solution
3
3
y(x) = c 1 e 3ln(x) + c 2 ln(x)e 3ln(x) = c 1 x + c 2 x ln(x)
for x > 0.
EXAMPLE 2.23
2
Solve x y + 3xy + 10y = 0.

The transformed equation is Y + 2Y + 10Y = 0 with the general solution

−t
−t
Y(t) = c 1 e cos(3t) + c 2 e sin(3t).
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October 14, 2010 14:12 THM/NEIL Page-73 27410_02_ch02_p43-76