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Yuliya Gorb

Reduction of Order.

Euler’s Equation

Week 6

September 26 – 30, 2016

Abel’s Theorem about Wronskian

We start with

Theorem (Abel)

Suppose that y 1 (t) and y 2 (t) are solutions of

y

′′

+ p(t)y

′

+ q(t)y = 0, (1)

with p(t) and q(t) that are continuous on some (a, b). Then the

Wronkian of y 1 and y 2 is given by

W y 1 , y 2 = Ce

−

∫ p(t)dt

where the constant C depends on solutions y 1 and y 2.

Note that since the Wronskian is a multiple of exponential

function, then it is either trivially zero (when C = 0), or some

non-zero function (when C 6 = 0)

Reduction of Order

We apply (2) to reduction of order problem. Suppose one is given a solution

y 1 (t) of ODE (1). The goal is to find a general solution of (1). For that we

need to seek for a second solution y 2 (t) that together with y 1 (t) forms a

fundamental set of solutions to (1). For that we seek for such y 2 that provides

the Wronskian (2) with const C = 1 together with the given y 1.

(^1) Then

W y 1 , y 2 =

y 1 (t) y 2 (t)

y

′ 1 (t)^ y^

′ 2 (t)

= y 1 (t)y

′ 2 (t)^ −^ y^

′ 1 (t)y^2 (t) =^ e

−

∫ p(t)dt ,

that is we obtain the 1st order linear ODE for the unknown y 2 that one

can solve using the integrating factor method (dropping constants). This

obtained solution y 2 does form a fundamental set of solutions to (1) since

we chose C = 1

(^2) Alternative approach: With the same as above we seek for y 2 that gives

C = 1 in (2) but assume that y 2 (t) = y 1 (t)v (t), where v (t) is TBD. Then

W y 1 , y 2 =

y 1 (t) y 1 (t)v (t)

y

′ 1 (t)^ y 1 (t)v^

′ (t) + y

′ 1 (t)v^ (t)

= y

2 1 (t)v^

′ (t) = e

−

∫ p(t)dt ,

or v

′ (t) = e

−

∫ p(t)dt /y

2 1 (t), which is solved by direct integration

Reduction of Order (cont.)

Example:

Given known solution y 1 (t) = t

− 1 find a general solution of

2 t

2 y

′′

• 3ty

′ − y = 0, t > 0

We already know that Abel’s Thm applied for this ODE yields:

W y 1 , y 2 = Ct

− 3 / 2

. Assume C = 1, then

(^1) W y 1 , y 2 = t

− 3 / 2

t

− 1 y 2

−t

− 2 y

′ 2

= t

− 1 y

′ 2 −^ t

− 2 y 2 = t

− 3 / 2

. Solve this

linear ODE y

′ − t

− 1 y 2 = t

− 1 / 2 for y 2 integrating factor is

μ(t) = e

−

∫ (^) dt t (^) = t

− 1 d

dt

[

t

− 1 y 2

]

= t

− 3 / 2 (drop a const)

y 2 (t) = t

1 / 2

. Then a general solution is y (t) = C 1 t

− 1

• C 2 t

1 / 2

(^2) Here we assume that y 2 (t) = t

− 1 v (t) and compute the corresponding

Wronskian: W y 1 , y 2 = t

− 3 / 2

t

− 1 t

− 1 v (t)

−t

− 2 t

− 1 v

′ (t) − t

− 2 v (t)

= t

− 2 v

′ (t). Obtain ODE for v (t): t

− 2 v

′ (t) = t

− 3 / 2 from which we find

v (t) = t

3 / 2 (by dropping one constant and keeping const 1 in front of

t

3 / 2 ). Then y 2 (t) = t

− 1 t

3 / 2 = t

1 / 2 and a general solution is as above

y (t) = C 1 t

− 1

• C 2 t

1 / 2

Euler’s Equation

We have three types of roots of (4), namely,

(^1) Roots of (4) are real and different: r 1 6 = r 2

(^2) Roots of (4) are complex conjugate: r 1 = λ + iμ, r 2 = λ − iμ

(^3) Roots of (4) are real and equal: r 1 = r 2 =: r

For simplicity we only consider the first case, when roots r 1 and r 2 of (4) are

real and different. Then the corresponding solutions of (3) are y 1 (t) = t

r 1 and

y 2 (t) = t

r 2

. Let’s check their Wronskian:

W [t

r 1 , t

r 2 ] =

t

r 1 t

r 2

r 1 t

r 1 − 1 r 2 t

r 2 − 1

= (r 2 − r 1 )t

r 1 +r 2 − 1 6 = 0 for t > 0 and if r 1 6 = r 2

So, y 1 (t) = t

r 1 and y 2 (t) = t

r 2 do form a fundamental a fundamental set of

solutions to (3) and its general solution is

y (t) = C 1 t

r 1

• C 2 t

r 2 , with r 1 , r 2 that solve (4)

Euler’s Equation (cont.)

Example:

Solve 2 t

2 y

′′

• 3ty

′ − y = 0, t > 0

It is an Euler equation and the roots of its characteristic equation

2 r (r − 1) + 3r − 1 = 2r

2

• r − 1 = 0 are r 1 = 1/2 and r 2 = −1 then the

general solution is

y (t) = C 1 t

1 / 2

• C 2 t

− 1