Answer to Question #196121 in Differential Equations for Ch.Trinadh

Question

Solve:(𝐷2 −2𝐷𝐷′ +𝐷′2)𝑍=𝑥2𝑦2𝑒𝑥+𝑦

Answer 1

$(D^2-2DD'+D'^2)z=x^2y^2e^x+y$

Auxiliary equation is given by

$m^2-2m+1=0$

$m=1,1$

Complimentary function is given by :

$u=\phi_1(y+1.x)+x\phi_2(y+1.x)$

$u=\phi_1(y+x)+x\phi_2(y+x)$

$D=\dfrac{d}{dx}\space\space\space\space\space\space\space\space\space\space D'=\dfrac{d}{dy}$

Particular integral, PI :

$=\dfrac{1}{(D^2-2DD'+D'^2)}x^2y^2e^x+\dfrac{1}{(D^2-2DD'+D'^2)}y$

$P.I_1=\dfrac{1}{(D^2-2DD'+D'^2)}e^xx^2y^2$

Operating the denominator on $e^x$

$=\dfrac{e^x}{((D+1)^2-2(D+1)D'+D'^2)}e^xx^2y^2$

$=e^x\dfrac{1}{[1+(D^2-2D-2D'-2DD'+D'^2)]}x^2y^2$

$=e^x[1+(D^2-2D-2D'-2DD'+D'^2)]^{-1}x^2y^2$

Using binomial expansion of $(1+x)^{-1}$, expanding the given equation

$=e^x[1+(D^2-2D-2D'-2DD'+D'^2)]x^2y^2$

Operating $D$ and $D'$ on $x^2y^2$

$=e^x[x^2y^2+2y^2+4xy^2-4x^2y-8xy+2x^2]$

$P.I_2=\dfrac{1}{(D^2-2DD'+D'^2)}y$

$=\dfrac{1}{(D-D')^2}y$

Taking $D^2$ common from the denominator

$=\dfrac{1}{D^2\bigg(1-\dfrac{D'}{D}\bigg)^2}y$

$=\dfrac{1}{D^2}\bigg(1+\dfrac{D'}{D}\bigg)^{-2}y$

Using binomial expansion of $(1+x)^{-2}$ and neglecting higher powers as they go to zero

$=\dfrac{1}{D^2}\bigg(1+\dfrac{2D'}{D}\bigg)y$

Operating $\bigg(1+\dfrac{2D'}{D}\bigg)$ on y

$=\dfrac{1}{D^2}y+\dfrac{2}{D^3}$

$=\int\int ydx+2\int\int\int dx$

$=\dfrac{yx^2}{2}+\dfrac{x^3}{6}$

Complete solution, $CS=u+PI_1+PI_2$

$z=\phi_1(y+x)+x\phi_2(y+x)+e^x[x^2y^2+2y^2+4xy^2-4x^2y-8xy+2x^2]+\dfrac{yx^2}{2}+\dfrac{x^3}{6}$