Answer to Question #272862 in Calculus for Lads

Question

perform this using the linear differential equation of higher order in operator form: (D2 + 3D + 2) (e-2x + 3x²)

Answer 1

Solution;

Given;

$(D^2+3D+2)y=e^{-2x}+3x^2$

The auxiliary equation is;

$m^2+3m+2=0$

$(m+2)(m+1)=0$

$m=-2 ,-1$

Therefore,the complementary solution is;

$C.F=c_1e^{-x}+c_2e^{-2x}$

The particular solutions are;

For $e^{-2x}$ ;

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

$P.I=\frac{e^{-2x}}{(-2+1)(D+2-2)}[1]$

$P.I=\frac{e^{-2x}}{(-1)}\frac1D{1}=-e^{-2x}x$

For $3x^2$;

$P.I=[\frac{1}{D^2+3D+2}]^{-1}3x^2$

$P.I=\frac12[1+\frac{(D^2+3D)}{2}+...]3x^2$

$P.I=\frac12[3x^2+3+9x]$

Hence;

$y(x)=C.F+P.I$

$y(x)=c_1e^{-x}+c_2e^{-2x}-xe^{-2x}+\frac32x^2+\frac92x+\frac32$