Linear Algebra and Differential Equations

Sujin Khomrutai, Ph.D.

Department of Math & Computer ScienceChulalongkorn University

Lecture 4

S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 1 / 14

Table of Contents

1 Cauchy-Euler Equations

S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 2 / 14

Table of Contents

1 Cauchy-Euler Equations

S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 3 / 14

Nonconstant coefficients equations

Now we study

an(t)y(n) + an−1(t)y

(n−1) + · · ·+ a0(t)y = 0

where ai (t) are continuous functions and an(t) 6= 0.

It can be extended to non-homogeneous equations.

Definition

If there are numbers αn, . . . , α1, α0 such that

an(t) = αntn, . . . , a1(t) = αi t, a0(t) = α0

the equation is called Cauchy-Euler equation.

Some αi may be zero except, so the term αi tiy (i) disappear.

S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 4 / 14

Nonconstant coefficients equations

EX. Consider ODEs

1 3t2y ′′ + 2ty ′ − 3y = 0 is a Cauchy-Euler equation.

2 t3y ′′′ − 5ty ′ + 3y = 0 is a Cauchy-Euler equation.

3 2t5y (5) − t4y (4) + t2y ′′ + 6y = 0 is a Cauchy-Euler equation.

4 ty ′′ − y ′ + (1/t)y = 0 seems not be a C-E in the first place.

Multiplying with t get

t2y ′′ − ty ′ + y = 0,

which is a Cauchy-Euler equation.

5 y ′′ + 2ty ′ − t2y = 0 is not a Cauchy-Euler equation.

S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 5 / 14

Cauchy-Euler Equations

Theorem

The transformationz = ln t, Y (z) = y(ez)

changes the Cauchy-Euler

at2y ′′ + bty ′ + cy = 0

(a, b, c are constants) into the form

aY ′′ + (b − a)Y ′ + cY = 0.

Backward transform.

t = ez , y(t) = Y (ln z).

S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 6 / 14

Cauchy-Euler equations

Proof. Since z = ln t we have

dz

dt= t−1 and

dt

dz= t.

By the chain rule, we calculate

y ′ =dy

dt=

dY

dz

dz

dt= t−1

dY

dz⇒ ty ′ = dY

dz.

Diff. z the last identity and apply chain rule:

dt

dzy ′ + ty ′′t =

d2Y

dz2⇒ t2y ′′ = d

2Y

dz2− dY

dz.

Plugging into the given ODE we obtain

a(Y ′′ − Y ′) + bY ′ + cY = 0 ⇒ aY ′′ + (b − a)Y ′ + cY = 0.

S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 7 / 14

Cauchy-Euler equation

EX. Solve the Cauchy-Euler equation

t2y ′′ + ty ′ + 4y = 0.

Sol. Let z = ln t and Y (z) = y(ez). By the theorem, the given ODE istransformed into

Y ′′ + (1− 1)Y ′ + 4Y = 0 ⇒ Y ′′ + 4Y = 0.

Solving the ODE we get

Y (z) = C1 cos 2z + C2 sin 2z .

Transform back: y(t) = Y (ln t) so

y(t) = C1 cos(2 ln t) + C2 sin(2 ln t).

S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 8 / 14

Cuachy-Euler equation

EX. Solve the ODE2t2y ′′ + 3ty ′ − y = 0.

Sol. Let z = ln t and Y (z) = y(ez). By the theorem, we get

2Y ′′ + (3− 2)Y ′ − Y = 0 ⇒ 2Y ′′ + Y ′ − Y = 0.

Characteristic equation: 2r2 + r − 1 = 0, r1 = −1, r2 = 1/2. So

Y (z) = C1e−z + C2e

z/2.

Transform back: y(t) = Y (ln t)

y(t) = C1t−1 + C2t

1/2.

S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 9 / 14

Cauchy-Euler equation

EX. Solve the ODEt2y ′′ − 5ty ′ + 9y = 0.

Sol. Let z = ln t and Y (z) = y(ez). Then the ODE becomes

Y ′′ − 6Y ′ + 9Y = 0.

Char. equation: r2 − 6r + 9 = 0, r1 = r2 = 3. So

Y (z) = (C1 + C2z)e3z .

Transform back: y(t) = Y (ln t) we get

y(t) = (C1 + C2 ln t)t3.

S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 10 / 14

Exercise

EX. Solve the ODEt2y ′′ + 8ty ′ + 12y = 0.

Sol. Let z = ln t and Y (z) = y(ez). The ODE becomes

Y ′′ + 7Y ′ + 12Y = 0.

Char. equation r2 + 7r + 12 = 0, r1 = −4, r2 = −3. SoY (z) = C1e

−4z + C2e−3z

Transform back: y(t) = Y (ln t) we get

y(t) = C1t−4 + C2t

−3.

S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 11 / 14

Nonhomogeneous Cauchy-Euler

We can employ the same transform

z = ln t, Y (z) = y(ez)

and the backward transform

t = ez , y(t) = Y (ln t).

to solve a nonhomogeneous Cauchy-Euler equation

at2y ′′ + bty ′ + cy = f (t).

Note the under the transform the ODE becomes

aY ′′ + (b − a)Y ′ + cY = f (ez).

S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 12 / 14

Nonhomogeneous Cauchy-Euler

EX. Solve the ODEt2y ′′ − 3ty ′ + 4y = ln t4.

Sol. Let z = ln t,Y (z) = y(ez). Then ODE becomes

Y ′′ − 4Y ′ + 4Y = 4 ln(ez) = 4z .

Complementary part. Y ′′ − 4Y ′ + 4Y = 0, get r1 = r2 = 2. So

Yc(z) = (C1 + C2z)e2z .

Particular sol. k = 0, non-resonance. So Yp(z) = Az + B, then getA = B = 1 hence Yp(z) = z + 1.

General sol. Y (z) = Yc(z) + Yp(z) = (C1 + C2z)e2z + z + 1.

Transform back: y(t) = Y (ln t)

y(t) = (C1 + C2 ln t)t2 + ln t + 1.

S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 13 / 14

Nonhomogeneous Cauchy-Euler

EX. Solve the ODEt2y ′′ − 2y = 3t2 − 1.

Sol. Let z = ln t,Y (z) = y(ez). Then ODE becomes

Y ′′ − Y ′ − 2Y = 3e2z − 1.

Complementary part. Y ′′ − Y ′ − 2Y = 0, get r1 = −1, r2 = 2. So

Yc(z) = C1e−z + C2e

2z .

Particular sol. Mixed, 3e2z (resonance, s = 1) and −1 (non-resonance).MUC: Yp(z) = Aze

2z + B, then get Yp(z) = ze2z + 12 .

General Sol. Y (z) = Yc(z) + Yp(z) = C1e−z + C2e2z + ze2z +12 .

Transform back: y(t) = Y (ln t)

y(t) = C1(1/t) + C2t2 + (ln t)t2 +

1

2.

S. Khomrutai (CU) Lin & Diff Eqns. Lecture 4 14 / 14

Cauchy-Euler Equations