Answer to Question #262855 in Differential Equations for harry

Question

solve 8y''-2y'-y=x²+x+1

Answer 1

Homogeneous differential equation is 8y"-2y'-y=0. -------(1)

Auxiliary equation,

8m²-2m-1=0

m=$\frac{1}{2},\frac{-1}{4}$

Therefore, solution of (1),

y(x)=$c_1e^{\frac{1}{2}x}+c_2e^{\frac{-1}{4}x}$

Now, finding the particular solution of the given differential equation.

$y_p=\frac{1}{8D^2-2D-1}(x^2+x+1)\\ =(-1)[1+(2D-8D^2)]^{-1}(x^2+x+1)\\ =(-1)[1-(2D-8D^2)+(2D-8D^2)^2-(2D-8D^2)^3-...](x^2+x+1)\\\text{Eliminate higher degree terms}\\ =(-1)[1-(2D-8D^2)+(2D-8D^2)^2](x^2+x+1)\\ =(-1)[x^2+x+1-(2x+1)+24]\\ =-x^2+3x-23\\ \text{Therefore, general solution is }\\ y(x)=c_1e^{\frac{1}{2}x}+c_2e^{\frac{-1}{4}x}-x^2+3x-23$