y" − (x + 1)y' — y = 0 a. do y = a₁ [1 + x² + ²x¹ + ² xº + ... ] + α₁ [x + ½ x³ + 1⁄2x5 + ²/3; +=x' + 1 1 b. y = ao [1 + ·x² + x³ + x² + -·|· + ... a1 x+ x += x² + + c. y = a₁ [1 + x² + x³ + ²x^² + ... C. + + d. y = ... · ao √x + ²/² x ² + ₁ x³ + [1⁄x² + ½ x³ + ¼ x4 + ···] ao ªo [⁄/ x² + ² x³ + ² x4 + ··· ] + a₁ + ·1
y" − (x + 1)y' — y = 0 a. do y = a₁ [1 + x² + ²x¹ + ² xº + ... ] + α₁ [x + ½ x³ + 1⁄2x5 + ²/3; +=x' + 1 1 b. y = ao [1 + ·x² + x³ + x² + -·|· + ... a1 x+ x += x² + + c. y = a₁ [1 + x² + x³ + ²x^² + ... C. + + d. y = ... · ao √x + ²/² x ² + ₁ x³ + [1⁄x² + ½ x³ + ¼ x4 + ···] ao ªo [⁄/ x² + ² x³ + ² x4 + ··· ] + a₁ + ·1
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
FOR REVIEW PURPOSES
Power Series
Find the Power series solution for the given differential equations centered at 0
![y" − (x + 1)y' — y = 0
a. do
y = a₁ [1 + x² + ²x¹ + ² xº + ... ] + α₁ [x + ½ x³ + 1⁄2x5 + ²/3;
+=x' + 1
1
b. y = ao [1 +
·x² +
x³
+ x² + -·|· +
...
a1
x+
x += x² +
+
c. y = a₁ [1 +
x² +
x³ + ²x^² + ...
C.
+
+
d. y = ...
· ao √x + ²/² x ² + ₁ x³ +
[1⁄x² + ½ x³ + ¼ x4 + ···]
ao
ªo [⁄/ x²
+ ² x³ + ² x4 + ··· ] + a₁
+
·1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F38c2618e-4acf-41c7-a022-94a725b363f8%2Fc480eda0-8864-4325-88ab-b19b9f0fa1d2%2F2plxvja_processed.png&w=3840&q=75)
Transcribed Image Text:y" − (x + 1)y' — y = 0
a. do
y = a₁ [1 + x² + ²x¹ + ² xº + ... ] + α₁ [x + ½ x³ + 1⁄2x5 + ²/3;
+=x' + 1
1
b. y = ao [1 +
·x² +
x³
+ x² + -·|· +
...
a1
x+
x += x² +
+
c. y = a₁ [1 +
x² +
x³ + ²x^² + ...
C.
+
+
d. y = ...
· ao √x + ²/² x ² + ₁ x³ +
[1⁄x² + ½ x³ + ¼ x4 + ···]
ao
ªo [⁄/ x²
+ ² x³ + ² x4 + ··· ] + a₁
+
·1
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