Answer to Question #199570 in Differential Equations for Rajkumar

Question

Solve the following PDEs:

i) (3D² - 2D² + D - 1 )z = 4e ^x+y cos(x + y )

ii) (D + D' - 1) (D + 2D' - 3) z = 4 + 3x + 6y .

Answer 1

$z=C.F.+P.I.$

I)

$D^2+D-1=(D+\frac{1+\sqrt{5}}{2})(D+\frac{1-\sqrt{5}}{2})$

$C.F.=e^{\frac{1+\sqrt{5}}{2}}\varphi_1(y)+e^{\frac{1-\sqrt{5}}{2}}\varphi_2(y)$

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

$=4e^{x+y}\frac{3D-2}{5}cos(x+y)=-\frac{4}{5}e^{x+y}(3sin(x+y)+2cos(x+y))$

ii)

$C.F.=e^x\varphi_1(y-x)+e^{3x}\varphi_2(y-x)$

$P.I.=\frac{1}{(D+D'-1)(D+2D'-3)}(4+3x+6y)=$

$=\frac{1}{3}(1+(D+D')+\frac{D+2D'}{3}+...)(4+3x+6y)=\frac{1}{3}(18+3x+6y)=$

$=6+x+2y$