Euler equations (a substitution for second-order DEs)
Here’s an example I did in class, for section 3.3. You have one similar HW problem from
that section, and I wanted to provide you with a good reference for how to do this.
This is the only type of change-of-variables substitution I would ask you about for
second-order DEs.
The problem. Consider the differential equation
t2
dy
d2 y
+ 2y = 0.
+
4t
dt2
dt
2
(I’m using the notation dy
instead of y 0 , and ddt2y for y 00 , because it’s clearer for this type of
dt
problem. It’s really helpful to remember what you’re differentiating with respect to.)
Note that we can’t use the characteristic equation technique to solve this DE, because
there are nonconstant functions of t being multiplied by y 00 and y 0 . We’re going to use the
substitution x = ln(t) to transform this DE into a DE with constant coefficients.
This same substitution and procedure will always work, as long as you have a DE of the
form at2 y 00 + bty 0 + cy = 0 (and these DEs are called “Euler equations”).
We’re going to try to get rid of all the t’s in this equation, leaving only y’s and x’s.
First step. Find a formula for
dy
,
dt
and then plug it into the DE.
By the chain rule, we can write
dy dx
dy
=
.
dt
dx dt
To see where this formula comes from, think of an actual example for y. For example, take
y = 2(ln(t))2 . Since x = ln(t), we could also write this as a function of x: y(x) = 2x2 . (This
might seem contrived because there was a ln(t) just hanging out, ready to be called x, but
you could plug ex in for t if this wasn’t the case.) Anyway, if you compute dy
, you should
dt
dy
−1
get 4 ln(t) · t , using the chain rule. Now compute the right side: dx = 4x, and dx
= t−1 . If
dt
you multiply these and remember that x = ln(t), you get exactly dy
. Do this on paper to be
dt
sure you understand.
The chain rule as I stated it above can look confusing, but it’s just a way of expressing
the above idea mathematically.
dy
Now we don’t know what y really is, so we can’t simplify dx
or dy
, but we definitely know
dt
1
what x is: x = ln(t), so dx
=
.
Therefore,
let’s
plug
it
in
and
make
things a bit nicer:
dt
t
dy
dy 1
=
· .
dt
dx t
Now plug this into the original DE in place of
dy
.
dt
Second step. Find a formula for the second derivative
d2 y
,
dt2
and plug it into the DE.
“But wait,” you say, “I already know the first derivative. Can’t I just differentiate it,
and be done?” To that, I say yes! That’s exactly what you need to do. But it might not be
immediately obvious how to differentiate, so I’d like to go through the steps.
Anyway, yes,
d2 y
dt2
is just the derivative (with respect to t) of
d2 y
d dy
=
dt2
dt dt
d dy 1
=
·
,
dt dx t
dy
.
dt
In other words,
dy
using our equation for dy
from above. Now dx
is some function of x (in our concrete example
dt
dy
a couple paragraphs up, dx = 4x, but we have no idea what it is in general). But x = ln(t),
dy
can be viewed as a function of t too: specifically, in the example, we’d have 4 ln(t),
so dx
but in general you don’t know the formula. But you do need to realize that the expression
inside the parentheses is actually a product of functions of t, so we use the product rule to
differentiate. The product rule says this:
d dy
1 dy d 1
d dy 1
·
=
· +
·
.
dt dx t
dt dx
t dx dt t
dy
The first expression in this sum is the hardest to think about. How to you get dtd dx
? You
need to use the chain rule again:
d2 y dx
d dy
= 2 .
dt dx
dx dt
dy
This looks way more confusing than it actually is, so I’d like to set dx
= f (x) so we can have
something that looks a little nicer. What if you wanted to differentiate f (x) with respect to
t, knowing that x was a function of t? You’d do this:
d
(f (x)) = f 0 (x) · x0 (t) or, using the other notation,
dt
df dx
=
· .
dx dt
2
dy
df
d y
But f was just dx
, so dx
= dx
2 , and you get the above formula. Remember that you can
1
dx
always plug in t for dt , since you know exactly what x is. And the second summand in the
dy 1
dy −1
expression for dtd dx
· t above is not bad: it’s dx
· t2 . (Let me know if you don’t see why.)
Page 2
Putting all the pieces together, we get
2
d2 y
1 dy −1
dy 1
· +
=
·
dt2
dx2 t
t dx t2
2
1 dy
dy
= 2
−
t dx2 dx
Now plug this too into the original DE.
Third step. Solve the new equation, with only y’s and x’s.
After you plug everything in to the original DE, you’ll get
d2 y
dy
+ 3 + 2y = 0.
2
dx
dx
(Check this for yourself!!!) The general solution is then y = c1 e−x + c2 e−2x . Plug x = ln(t)
back in to get the general solution in terms of t:
y = c1 t−1 + c2 t−2 .
Page 3