- 2071 Shrawan · 5 marks
Solve y''−2y'+y=x²e^(3x)
Short answer
The complementary function is (c₁ + c₂x)eˣ. Using the shift rule, the particular integral is (e^(3x)/4)(x² - 2x + 3/2), so y = (c₁ + c₂x)eˣ + (e^(3x)/4)(x² - 2x + 3/2).
Solve: d²y/dx² − 2 dy/dx + y = x²e^(3x).
The equation is (D² − 2D + 1)y = x²e^(3x).
Complementary function
m² − 2m + 1 = 0, so m = 1, 1.
CF = (c₁ + c₂x)eˣ
Particular integral
Replace D by D + 3 and take e^(3x) outside:
PI = e^(3x) · 1/(D + 3 − 1)² · x² = e^(3x) · 1/(D + 2)² · x² = e^(3x) · (1/4)(1 + D/2)⁻² x² = (e^(3x)/4)(1 − D + (3/4)D² − …) x² = (e^(3x)/4)(x² − 2x + 3/2)
General solution
y = (c₁ + c₂x)eˣ + (e^(3x)/4)(x² − 2x + 3/2)
✓ Answer checked by Claude Opus 5.5 (high effort)
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