Give the complete solution of this general linear homogeneous functio

1.	Give the complete solution of this general linear homogeneous function.y'''- y'' = e3x + 3e4x
Answer» Given differential equation is y'''-y''=e^3x+3 e^4x It's auxiliary equation is m³-m² = 0 = m²(m-1) = 0 = m = 0, 0 & 1. ∴ CF = (C₁ + C₂x)e0x + C₃e^x = C₁ + C₂x + C₃e^x P.I. = \(\frac {1}{D^3-D^2} (e^{3x} + 3e^{4x)}\) = \(\frac {1}{D^2(D-1)}e^{3x} + \frac {3}{D^2(D-1)}e^{4x}\) = \(\frac {e^{3x}}{3^2(3-1)} + \frac {3e^{4x}}{4^2(4-1)}\) = \(\frac {e^{3x}}{18} + \frac {3e^{4x}}{48}\) = \(\frac {e^{3x}}{18} + \frac {e^{4x}}{16}\) ∴ Complete solution of given differential equation is y = CF + P.I. = C₁ + C₂x + C₃e^x + \(\frac {e^{3x}}{18} + \frac {e^{4x}}{16}\)

Give the complete solution of this general linear homogeneous function.y'''- y'' = e3x + 3e4x

Answer»

Given differential equation is y'''-y''=e^3x+3 e^4x

It's auxiliary equation is

m³-m² = 0

= m²(m-1) = 0

= m = 0, 0 & 1.

∴ CF = (C₁ + C₂x)e0x + C₃e^x

= C₁ + C₂x + C₃e^x

P.I. = \(\frac {1}{D^3-D^2} (e^{3x} + 3e^{4x)}\)

= \(\frac {1}{D^2(D-1)}e^{3x} + \frac {3}{D^2(D-1)}e^{4x}\)

= \(\frac {e^{3x}}{3^2(3-1)} + \frac {3e^{4x}}{4^2(4-1)}\)

= \(\frac {e^{3x}}{18} + \frac {3e^{4x}}{48}\)

= \(\frac {e^{3x}}{18} + \frac {e^{4x}}{16}\)

∴ Complete solution of given differential equation is

y = CF + P.I.

= C₁ + C₂x + C₃e^x + \(\frac {e^{3x}}{18} + \frac {e^{4x}}{16}\)

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