Linear Algebra and Differential Equations

Linear Algebra and Differential Equations

Sujin Khomrutai, Ph.D.

Department of Math & Computer Science Chulalongkorn University

Lecture 4

Table of Contents

1 Cauchy-Euler Equations

Table of Contents

1 Cauchy-Euler Equations

Nonconstant coefficients equations

Now we study

a_n(t)y⁽ⁿ⁾+an−1(t)y⁽ⁿ⁻¹⁾+· · ·+a₀(t)y= 0 where a_i(t) are continuous functions anda_n(t)6= 0.

It can be extended to non-homogeneous equations.

Definition

If there are numbersαn, . . . , α1, α0 such that

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

the equation is called Cauchy-Euler equation.

Someαi may be zero except, so the term αitⁱy⁽ⁱ⁾ disappear.

Nonconstant coefficients equations

EX.Consider ODEs

1 3t²y⁰⁰+ 2ty⁰−3y = 0 is a Cauchy-Euler equation.

2 t³y⁰⁰⁰−5ty⁰+ 3y = 0 is a Cauchy-Euler equation.

3 2t⁵y⁽⁵⁾−t⁴y⁽⁴⁾+t²y⁰⁰+ 6y= 0 is a Cauchy-Euler equation.

4 ty⁰⁰−y⁰+ (1/t)y = 0 seems not be a C-E in the first place.

Multiplying witht get

t²y⁰⁰−ty⁰+y= 0, which is a Cauchy-Euler equation.

5 y⁰⁰+ 2ty⁰−t²y = 0 is not a Cauchy-Euler equation.

Cauchy-Euler Equations

Theorem

The transformation

z = lnt, Y(z) =y(e^z) changes the Cauchy-Euler

at²y⁰⁰+bty⁰+cy = 0 (a,b,c are constants) into the form

aY⁰⁰+ (b−a)Y⁰+cY = 0.

Backward transform.

t =e^z, y(t) =Y(lnz).

Cauchy-Euler equations

Proof. Since z = lnt we have dz

dt =t⁻¹ and dt dz =t. By the chain rule, we calculate

y⁰= dy dt = dY

dz dz

dt =t⁻¹dY

dz ⇒ ty⁰= dY dz. Diff. z the last identity and apply chain rule:

dt

dzy⁰+ty⁰⁰t= d²Y

dz² ⇒ t²y⁰⁰= d²Y dz² −dY

dz. Plugging into the given ODE we obtain

a(Y⁰⁰−Y⁰) +bY⁰+cY = 0 ⇒ aY⁰⁰+ (b−a)Y⁰+cY = 0.

Cauchy-Euler equation

EX.Solve the Cauchy-Euler equation

t²y⁰⁰+ty⁰+ 4y = 0.

Sol. Letz = lnt and Y(z) =y(e^z). By the theorem, the given ODE is transformed into

Y⁰⁰+ (1−1)Y⁰+ 4Y = 0 ⇒ Y⁰⁰+ 4Y = 0.

Solving the ODE we get

Y(z) =C1cos 2z+C2sin 2z.

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

y(t) =C₁cos(2 lnt) +C₂sin(2 lnt).

Cuachy-Euler equation

EX.Solve the ODE

2t²y⁰⁰+ 3ty⁰−y = 0.

Sol. Letz = lnt and Y(z) =y(e^z). By the theorem, we get 2Y⁰⁰+ (3−2)Y⁰−Y = 0 ⇒ 2Y⁰⁰+Y⁰−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(lnt)

y(t) =C₁t⁻¹+C₂t^1/2.

Cauchy-Euler equation

EX.Solve the ODE

t²y⁰⁰−5ty⁰+ 9y = 0.

Sol. Letz = lnt and Y(z) =y(e^z). Then the ODE becomes Y⁰⁰−6Y⁰+ 9Y = 0.

Char. equation: r²−6r+ 9 = 0, r1 =r2 = 3. So Y(z) = (C1+C2z)e^3z. Transform back: y(t) =Y(lnt) we get

y(t) = (C₁+C₂lnt)t³.

Exercise

EX.Solve the ODE

t²y⁰⁰+ 8ty⁰+ 12y= 0.

Sol. Letz = lnt and Y(z) =y(e^z). The ODE becomes Y⁰⁰+ 7Y⁰+ 12Y = 0.

Char. equation r²+ 7r+ 12 = 0, r1 =−4,r2 =−3. So Y(z) =C1e^−4z+C2e^−3z Transform back: y(t) =Y(lnt) we get

y(t) =C₁t⁻⁴+C₂t⁻³.

Nonhomogeneous Cauchy-Euler

We can employ the same transform

z = lnt, Y(z) =y(e^z) and the backward transform

t =e^z, y(t) =Y(lnt).

to solve a nonhomogeneous Cauchy-Euler equation at²y⁰⁰+bty⁰+cy =f(t).

Note the under the transform the ODE becomes aY⁰⁰+ (b−a)Y⁰+cY =f(e^z).

Nonhomogeneous Cauchy-Euler

EX.Solve the ODE

t²y⁰⁰−3ty⁰+ 4y = lnt⁴. Sol. Letz = lnt,Y(z) =y(e^z). Then ODE becomes

Y⁰⁰−4Y⁰+ 4Y = 4 ln(e^z) = 4z.

Complementary part. Y⁰⁰−4Y⁰+ 4Y = 0, getr1=r2 = 2. So Y_c(z) = (C₁+C₂z)e^2z.

Particular sol. k = 0, non-resonance. SoY_p(z) =Az+B, then get A=B = 1 henceYp(z) =z+ 1.

General sol. Y(z) =Yc(z) +Yp(z) = (C1+C2z)e^2z+z+ 1.

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

y(t) = (C1+C2lnt)t²+ lnt+ 1.

Nonhomogeneous Cauchy-Euler

EX.Solve the ODE

t²y⁰⁰−2y = 3t²−1.

Sol. Letz = lnt,Y(z) =y(e^z). Then ODE becomes Y⁰⁰−Y⁰−2Y = 3e^2z−1.

Complementary part. Y⁰⁰−Y⁰−2Y = 0, get r1 =−1,r2 = 2. So Y_c(z) =C₁e^−z+C₂e^2z.

Particular sol. Mixed, 3e^2z (resonance,s = 1) and−1 (non-resonance).

MUC:Yp(z) =Aze^2z+B, then getYp(z) =ze^2z+ ¹₂.

General Sol. Y(z) =Y_c(z) +Y_p(z) =C₁e^−z+C₂e^2z+ze^2z+¹₂. Transform back: y(t) =Y(lnt)

y(t) =C1(1/t) +C2t²+ (lnt)t²+ 1 2.