Substitute this int the differential Equation: y" - 8y = 16x2 (6 Ax + 2B) – 8 (Ax² + Bx +c) = 16m² Equating coefficients of like terms: - 8A2 + (64-88) x + (28-8c)=16x2 Equating Coefficients: -8A = 16 =) A = −2 6A -8B0 =) - 12-8B = 0 =) 6 = -12 = -3/2 So the particular Solution is: 3. Geneal Jp(x) = – 2x – 3/2 × Solution Sum of Solutions. IS Κ -3/8 If the homogeneous and particle general solution y(x) = Y₁ (x) + Yp(x) →) y(x) = Ce² + 2 x cos (√34) + Cze din √3x) 2x² -¾½ x – 38.
Substitute this int the differential Equation: y" - 8y = 16x2 (6 Ax + 2B) – 8 (Ax² + Bx +c) = 16m² Equating coefficients of like terms: - 8A2 + (64-88) x + (28-8c)=16x2 Equating Coefficients: -8A = 16 =) A = −2 6A -8B0 =) - 12-8B = 0 =) 6 = -12 = -3/2 So the particular Solution is: 3. Geneal Jp(x) = – 2x – 3/2 × Solution Sum of Solutions. IS Κ -3/8 If the homogeneous and particle general solution y(x) = Y₁ (x) + Yp(x) →) y(x) = Ce² + 2 x cos (√34) + Cze din √3x) 2x² -¾½ x – 38.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:Substitute
this
int
the differential
Equation:
y" - 8y
= 16x2
(6 Ax + 2B) – 8 (Ax² + Bx +c) = 16m²
Equating coefficients of like terms:
- 8A2 + (64-88) x + (28-8c)=16x2
Equating Coefficients:
-8A
= 16
=) A = −2
6A -8B0 =) - 12-8B = 0
=) 6 = -12 = -3/2
So the particular Solution is:
3. Geneal
Jp(x) = – 2x – 3/2 ×
Solution
Sum of
Solutions.
IS
Κ
-3/8
If the homogeneous and particle
general solution
y(x) = Y₁ (x) +
Yp(x)
→) y(x) = Ce² + 2 x cos (√34) + Cze din √3x)
2x² -¾½ x – 38.
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