Answer to Question #283472 in Differential Equations for Ruchi

Question

Solve (D²-2DD')=x³y+e^5x

Answer 1

$(D^2-2DD')z=x^3y+e^{5x}$

The auxillary equation is $m^2-2m=0,\quad m(m-2)=0,\quad m_1=0,\ m_2=2.$

Complementary function is $F=f_1(y)+f_2(y+2x)$

1 particular integral is $\frac{1}{D^2-2DD'}e^{5x+0y}=\frac{1}{5^2-2\cdot 5\cdot 0}=\frac{1}{25}e^{5x}$

2 particular integral is $\frac{1}{D^2-2DD'}x^3y=\frac{1}{D^2}\cdot \frac{1}{1-\tfrac{2D'}{D}}x^3y=\frac{1}{D^2}\big(1+\frac{2D'}{D}+\big(\frac{2D'}{D}\big)^2+...\big)x^3y=\frac{1}{D^2}(x^3y+\frac{2}{D}x^3+\frac{4}{D^2}0+...)=\frac{1}{D^2}x^3y+\frac{2}{D^3}x^3=\tfrac{1}{4\cdot 5}x^5y+\frac{2}{4\cdot 5\cdot 6}x^6=\tfrac{1}{20}x^5y+\tfrac{1}{60}x^6$

Answer: $z(x,y)=f_1(y)+f_2(y+2x)+\tfrac{1}{25}e^{5x}+\tfrac{1}{20}x^5y+\tfrac{1}{60}x^6.$