Answer to Question #293598 in Differential Equations for Saravana

Question

(D² − 3DD + 2D²)z = e²x+³y +sin(x-2y)

Answer 1

The given question is;

$\displaystyle (D^2 − 3DD\prime + 2(D\prime)^2)z = e^{2x+3y}+\sin(x-2y)$

Now put $D=m,\text{and }D\prime=1$ then the auxiliary equation is;

$\displaystyle m^2-3m+2=0\Rightarrow(m-2)(m-1)=0\Rightarrow m=1,2$

Hence, the complementary function is;

$\displaystyle z_c=f_1(y+x)+f_2(y+2x)$.

Next, for the particular integral ($\displaystyle y_p$ );

$\displaystyle z_p=\frac{1}{D^2 − 3DD\prime + 2(D\prime)^2}(e^{2x+3y}+\sin(x-2y))\\ \quad=\frac{1}{D^2 − 3DD\prime + 2(D\prime)^2}e^{2x+3y}+\frac{1}{D^2 − 3DD\prime + 2(D\prime)^2}\sin(x-2y)\\ \quad=\frac{1}{2^2-3(2)(3)+2(3)^2}e^{2x+3y}+\frac{1}{-(1)^2 − 3[-1\times(-2)] + 2[-(-2)^2]}\sin(x-2y)\\ \quad=\frac{1}{4-18+18}e^{2x+3y}+\left(\frac{1}{-1-3[2]+2[-4]}\sin(x-2y)\right)\\ \quad=\frac{1}{4}e^{2x+3y}+\frac{1}{(-1-6-8)}\sin(x-2y)\\ \quad=\frac{1}{4}e^{2x+3y}+\frac{1}{(-15)}\sin(x-2y)\\ \quad=\frac{1}{4}e^{2x+3y}-\frac{1}{15}\sin(x-2y)$

Thus the general solution is ; $\displaystyle z=z_c+z_p=f_1(y+x)+f_2(y+2x)+\frac{1}{4}e^{2x+3y}-\frac{1}{15}\sin(x-2y)$