Lecture Notes in Differential Equations - Bruce E. Shapiro

Lecture Notes in Differential Equations - Bruce E. Shapiro Lecture Notes in Differential Equations - Bruce E. Shapiro

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278 LESSON 29. REGULAR SINGULARITIES for some unknown number (integer, real, or possibly complex) r. To see what values of r might work, we substitute into the original ODE (29.4): t 2 r(r − 1)t r−2 + αtrt r−1 + βt r = 0 (29.16) =⇒ t r [r(r − 1) + αr + β] = 0 (29.17) Since this must hold for all values of t, the second factor must be identically equal to zero. Thus we obtain r(r − 1) + αr + β = 0 (29.18) without the restriction that r ∈ Z. We call this the indicial equation (even though there are no indices) due to its relationship (and similarity) to the indicial equation that we will have to solve in Frobenius’ method in chapter 30. We will now consider each of the three possible cases for the roots r 1,2 = 1 − α ± √ (1 − α) 2 − 4β 2 (29.19) Case 1: Two real roots. If r 1 , r 2 ∈ R and r 1 ≠ r 2 , then we have found two linearly independent solutions and the general solution is Example 29.4. Find the general solution of In standard from this becomes The indicial equation is y = C 1 t r1 + C 2 t r2 (29.20) 2t 2 y ′′ + 3ty ′ − y = 0 (29.21) t 2 y ′′ + 3 2 ty′ − 1 2 y = 0 (29.22) 0 = r(r − 1) − 3 2 r + 1 2 = 2 2 r2 + 1 2 r + 1 2 (29.23) 0 = 2r 2 + r + 1 = (2r − 1)(r + 1) =⇒ r = 1 , −1 (29.24) 2 Hence y = C 1 √ t + C 2 t (29.25)

279 Case 2: Two equal real roots. Suppose r 1 = r 2 = r. Then r = 1 − α 2 (29.26) because the square root must be zero (ie., (1 − α) 2 − 4β = 0) to give a repeated root. We have one solution is given by y 1 = t r = t (1−α)/2 . To get a second solution we can use reduction of order. From Abel’s formula the Wronskian is ( ∫ ) αtdt W = C exp − t 2 = C exp ( −α ∫ ) dt = Ce −α ln t = C t t α (29.27) By the definition of the Wronskian a second formula is given by W = y 1 y 2 ′ − y 2 y 1 ′ (29.28) ( = t (1−α)/2) ( ) 1 − α ( y 2 ′ − y 2 ′ t (−1−α)/2) (29.29) 2 Equating the two expressions for W , ( t (1−α)/2) ( ) 1 − α ( y 2 ′ − y 2 ′ t (−1−α)/2) = C 2 t α (29.30) ( ) 1 − α t y 2 ′ (−1−α)/2 C − 2 t (1−α)/2 y′ 2 = (29.31) t α (t (1−α)/2 ) ( ) α − 1 y 2 ′ + y ′ C 2 = (29.32) 2t t (1+α)/2 This is a first order linear equation; an integrating factor is (∫ ) ( ) α − 1 α − 1 µ(t) = exp dt = exp ln t = t (α−1)/2 (29.33) 2t 2 If we denote q(t) = Ct −(1+α)/2 then the general solution of (29.32) is y = 1 [∫ ] µ(t)q(t)dt + C 1 (29.34) µt [∫ ] = t (1−α)/2 t (α−1)/2 Ct −(1+α)/2 dt + C 1 (29.35) ∫ dt = Cy 1 t + C 1y 1 (29.36) = Cy 1 ln |t| + C 1 y 1 (29.37)

278 LESSON 29. REGULAR SINGULARITIES<br />

for some unknown number (<strong>in</strong>teger, real, or possibly complex) r. To see<br />

what values of r might work, we substitute <strong>in</strong>to the orig<strong>in</strong>al ODE (29.4):<br />

t 2 r(r − 1)t r−2 + αtrt r−1 + βt r = 0 (29.16)<br />

=⇒ t r [r(r − 1) + αr + β] = 0 (29.17)<br />

S<strong>in</strong>ce this must hold for all values of t, the second factor must be identically<br />

equal to zero. Thus we obta<strong>in</strong><br />

r(r − 1) + αr + β = 0 (29.18)<br />

without the restriction that r ∈ Z. We call this the <strong>in</strong>dicial equation<br />

(even though there are no <strong>in</strong>dices) due to its relationship (and similarity)<br />

to the <strong>in</strong>dicial equation that we will have to solve <strong>in</strong> Frobenius’ method <strong>in</strong><br />

chapter 30.<br />

We will now consider each of the three possible cases for the roots<br />

r 1,2 = 1 − α ± √ (1 − α) 2 − 4β<br />

2<br />

(29.19)<br />

Case 1: Two real roots. If r 1 , r 2 ∈ R and r 1 ≠ r 2 , then we have found two<br />

l<strong>in</strong>early <strong>in</strong>dependent solutions and the general solution is<br />

Example 29.4. F<strong>in</strong>d the general solution of<br />

In standard from this becomes<br />

The <strong>in</strong>dicial equation is<br />

y = C 1 t r1 + C 2 t r2 (29.20)<br />

2t 2 y ′′ + 3ty ′ − y = 0 (29.21)<br />

t 2 y ′′ + 3 2 ty′ − 1 2 y = 0 (29.22)<br />

0 = r(r − 1) − 3 2 r + 1 2 = 2 2 r2 + 1 2 r + 1 2<br />

(29.23)<br />

0 = 2r 2 + r + 1 = (2r − 1)(r + 1) =⇒ r = 1 , −1 (29.24)<br />

2<br />

Hence<br />

y = C 1<br />

√<br />

t +<br />

C 2<br />

t<br />

(29.25)

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