Question

(D^ 2 -40D' +4D'^2)2=e^ (2x+y)

Answer 1

The actual question is;

$\displaystyle (D^2-4DD^\prime+4(D^\prime)^2)z=e^{2x+y}, \text{where }D=\frac{\partial}{\partial x}, \text{and}\ D^\prime =\frac{\partial}{\partial y}$

To find the complementary function (C.F), let $\displaystyle D=m, \text{and }D^\prime=1,$ then the auxiliary equation of the given PDE in terms of variable m is;

$\displaystyle m^2-4m+4=0$

$\displaystyle \Rightarrow (m-2)(m-2)=0\Rightarrow m=2(\text{twice})$

Thus the complementary function of the given PDE is;

$\displaystyle \color{green}{C.F=z_c=f_1(y+2x)+xf_2(y+2x)}$

Next, to find the particular integral (P.I) of the given PDE, we proceed as follows;

$\displaystyle P.I=\frac{1}{(D^2-4DD^\prime+4(D^\prime)^2)}e^{2x+y}=x\frac{1}{(2D-4D^\prime)}e^{2x+y}=\frac{x^2}{2}e^{2x+y}.$

Thus the particular integral of the given PDE is;

$\displaystyle \color{blue}{P.I=z_p=\frac{x^2}{2}e^{2x+y}}$

Since complete solution (z) is $\displaystyle \color{red}{z=z_c+z_p}$, thus we have the complete solution as;

$\displaystyle \color{red}{z=f_1(y+2x)+xf_2(y+2x)+\frac{x^2}{2}e^{2x+y}}$