Answer to Question #237598 in Differential Equations for Tonika

Question

(D

3 − 3DD

′2 − 2D

′3

)z = cos(x + 2y)

Answer 1

(D³-3DD'-2D'³)z=Cos(x+2y)

$\because$ D=m and D'=1

The auxiliary equation is;

m³-3m-2=0

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

m=-1,-1,2

Therefore C.F=f₁(y-x)+xf₂(y-x)+f₃(y+2x)

Now,

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

But D²=-1,D'²=-2²=-4

P.I=$\frac{1}{(-1)D-3D(-4)-2(-4)D'}Cos(x+2y)$

=$\frac{1}{-D+12D+8D'}Cos(x+2y)$

= $\frac{1}{11D+8D'}Cos(x+2y)$

=$\frac{D}{11D^{2}+2DD'}Cos(x+2y)$

=$\frac{D}{11(-1)+8(-2)}Cos(x+2y)\ \because D^{2}=-1^{2}\ and\ D-D'=-1*2=-2$

P.I=$\frac{D}{-11-16}\int Cos(x+2y)$

=$\frac{-1}{27}[-Sin(x+2y)]$

=$\frac{1}{27}Sin(x+2y)$

The general solution becomes;

y=C.F+P.I

=f₁(y-x)+xf₂(y-x)+f₃(y+2x)+$\frac{1}{27}Sin(x+2y)$