Euler Equations

R. C. Daileda

A second order homogeneous linear ODE of the form

t2d2y

dt2+ αt

dy

dt+ βy = 0, (1)

with α, β some constants, is called an Euler equation. In this note we will show how a simple change ofvariables (analogous to a substitution in an integral) can be used to transform an Euler equation into a secondorder homogeneous linear ODE with constant coefficients. Since we know how to find the general solutionto the latter type of equation, we can therefore always find the general solution to any Euler equation.

We begin with the change of variablex = ln t.

Since the set of real numbers x corresponds in a one-to-one fashion with the set of positive numbers t, wecan view any function y of t as a function of x, and vice-versa. For example, the function of t

y = 12t− sin t

becomes the function of xy = 12ex − sin(ex)

since t = ex.Our immediate task is to determine how changing the independent variable from t to x changes the given

differential equation. The chain rule tells us how the derivative of y with respect to the “new” variable x isrelated to the derivative of y with respect to the “old” variable t. Indeed:

dy

dx=

dy

dt

dt

dx=

dy

dtt. (2)

Here we have used the fact that t = ex so that dt/dx = ex = t. Likewise, we can use the chain rule again(together with the product rule) to get an analogous expression for the second derivative of y with respectto x. Using the above we have

d2y

dx2=

d

dx

dy

dx

=d

dx

(dy

dtt

)=

(d

dx

dy

dt

)t +

dy

dt

dt

dx

=d2y

dt2dt

dxt +

dy

dt

dt

dx

=d2y

dt2t2 +

dy

dtt

where we have again used the fact that dt/dx = t. Combining this with (2) gives the relations

dy

dtt =

dy

dx(3)

dy

dtt2 =

d2y

dx2− dy

dx. (4)

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We now substitute (3) and (4) into the Euler equation (1) to see what it becomes as a result of our changeof variables. We get (

d2y

dx2− dy

dx

)+ α

dy

dx+ βy = 0.

Putting the dy/dx terms together, we are led to the following.

Conclusion. Under the change of variables x = ln t, the Euler equation

t2d2y

dt2+ αt

dy

dt+ βy = 0

is transformed into the constant-coefficient equation

d2y

dx2+ (α− 1)

dy

dx+ βy = 0. (5)

The moral of the story is that we now have a procedure for finding the general solution to Euler equations:

1. Perform the change of variable x = ln t, yielding the constant coefficient equation (5).

2. Find the general solution to (5) using the techniques we already know. This solution will be in termsof the independent variable x.

3. Eliminate x by substituting x = ln t, yielding the general solution to (1) in terms of the originalindependent variable t.

We demonstrate this procedure with a few examples.

Example 1. Find the general solution to the Euler equation

t2y′′ − 3ty′ + 7y = 0, t > 0. (6)

Solution. According to the work above, the variable change x = ln t transforms the given differentialequation into

d2y

dx2− 4

dy

dx+ 7y = 0 (7)

(we have α = −3, β = 7). The characteristic equation of this DE is

r2 − 4r + 7 = 0

which has the roots

r =4±

√16− 282

= 2±√

3i.

Therefore, the general solution to (7) is

y = c1e2x cos

√3x + c2e

2x sin√

3x.

To get the solution to (6) we simply remember the relationship x = ln t:

y = c1t2 cos(

√3 ln t) + c2t

2 sin(√

3 ln t).

Example 2. Find the general solution to the equation

4t2y′′ + 8ty′ − 3y = 0, t > 0. (8)

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Solution. Once we divide (8) by 4, we recognize this as an Euler equation with α = 2 and β = −3/4.Consequently, we know that the variable substitution x = ln t transforms (8) into the equivalent equation

d2y

dx2+

dy

dx− 3

4y = 0. (9)

The characteristic equation of the latter is

r2 + r − 34

=(

r +32

) (r − 1

2

)= 0

which has the roots r1 = 1/2 and r2 = −3/2. Therefore, the general solution to (9) is

y = c1ex/2 + c2e

−3x/2.

Switching back to the variable t through our substitution x = ln t we are led to the general solution of (8):

y = c1t1/2 + c2t

−3/2.

Example 3. Find the general solution to the equation

t2y′′ − 2y = 0. (10)

Solution. Notice that we considered this example in class: we found the general solution after an“oracle” had given us two fundamental solutions. Now we can find these fundamental solutions ourselves!

The equation (10) is an Euler equation with α = 0 and β = −2. The usual change of variables x = ln ttherefore turns it into

d2y

dx2− dy

dx− 2y = 0 (11)

which has characteristic equationr2 − r − 2 = (r − 2)(r + 1) = 0

so that the general solution isy = c1e

2x + c2e−x.

Changing back to t via x = ln t we find that the general solution to (11) is

y = c1t2 + c2t

−1.

This is exactly what we showed in class!

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