View
13
Download
0
Category
Preview:
Citation preview
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
Recommended