Question

(D³-12D+16)y=(e^x +e^-2x)²

Answer 1

(D³ - 12D + 16)y = (e^x + e^-2x)²

It's auxiliary equation is

m³ - 12m + 16 = 0

⇒ m²(m - 2) + 2m(m - 2) - 8 (m - 2) = 0

⇒ (m - 2) (m² + 2m - 8) = 0

⇒ m = 2, 2, -4

\(\therefore\) C.F. = (C₁ + C₂x)e^2x + C₃e^-4x

P.I. = \(\frac1{D^3-12D+16}(e^x+e^{-2x})^2\)

 = \(\frac1{D^3-12D+16}(e^{2x}+e^{-4x}+2e^{-x})\) 

= \(\frac1{D^3-12D+16}e^{2x}\) + \(\frac1{D^3-12D+16}e^{-4x}\) + \(\frac1{D^3-12D+16}e^{-x}\) 

= \(\frac1{(D-2)^2(D+4)}e^{2x}+\frac{1}{(D-2)^2(D+4)}e^{-4x}\) + \(\frac{2e^{-x}}{(-1)^3-12(-1)+16}\) 

 = \(\frac{x^2e^{2x}}{(2+4)\times2!}+\frac{xe^{-4x}}{(-4-2)^2\times1!}\) + \(\frac2{27}e^{-x}\)

\(\therefore\) Complete solution of given differential equation is

y = C.F. + P. I.

 = (C₁ + C₂x)e^2x + C₃e^-4x + \(\frac1{12}\)x²e^2x + \(\frac1{36}\)xe^-4x + \(\frac{2}{27}\)e^-x