Answer to Question #250487 in Differential Equations for KAMI

Question

Solve ( D3 – 3DD′2+2D′3 )Z = e2x – y

Answer 1

$( D^3 – 3DD'^2+2D'^3)Z = e^{2x – y}$

Replace D with m and D' with 1

Auxiliary equation becomes $m^3-3m+2=0$

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

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

m=1,1,-2

$C.F=f_1(y+x)+xf_2(y+x)+f_3(y-2x)$

For particular integral;

$P.I=\frac{1}{( D^3 – 3DD'^2+2D'^3)} e^{2x – y}$

Put D=2 and D'=-1

$P.I=\frac{1}{( 2^3 – 3(2)(-1)^2+2(-1)^3)} e^{2x – y}$

$=\frac{1}{8-6-2)}e^{2x-y}$

We then differentiate the denominator w.r.t D and multiply by x to get,

$P.I=\frac{x}{\frac{d}{dD}( D^3 – 3DD'^2+2D'^3)} e^{2x – y}$

$=\frac{x}{( 3D^2 – 3D'^2+0)} e^{2x – y}$

$=x\frac{1}{( 3D^2 – 3D'^2)} e^{2x – y}$

Now put D=2 and D'=-1

$=x\frac{1}{( 3(2)^2 – 3(-1)^2)} e^{2x – y}\newline=x\frac{1}{( 12– 3)} e^{2x – y}$

$P.I=\frac{x}{9}e^{2x-y}$

The solution to the equation becomes,

z=C.F+P.I

$=f_1(y+x)+xf_2(y+x)+f_3(y-2x)+\frac{1}{9}xe^{2x-y}$