Answer to Question #250485 in Differential Equations for KAMI

Question

Solve ( D3 – 3DD′2+2D′3 )Z = e2x – y

Answer 1

Given $(D^{3}-3DD'^{2*}+2D'^{3})z=e^{2x-y}$

Put D=m and D'=1

Auxiliary equation is;

m³-3m+2=0

(m-1)(m²+m-2)=0

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

m=1, 1, -2

C.F=f₁(y+x)+xf₂(y+x)+f₃(y-2x)

For particular integral,

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

Put D=2, D'=-1

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

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

On putting D=2 and D'=-1 in the denominator,it became zero.

Since the denominator becomes zero, we differentiate it with respect to D and multiply by x, we 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 out D=2 ,D'=-1

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

$=x\frac{1}{12-3}e^{2x-y}$

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

So the solution becomes,

z=C.F+P.I

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