Answer to Question #199687 in Differential Equations for anuj

Question

(D²-D'²-3D+3D')z=xy+7

Answer 1

Rewriting the given equation :

$(D-D')(D+D'-3)=xy+7$

comparing this with standard equation

$(D-m_1D'-\alpha_1)(D-m_2D'-\alpha_2)z=F(x,y)$

$m_1 = 1;\space\space\space\space\space\space\space\space m_2 = – 1 \\ \alpha_1 = 0; \space\space\space\space\space\space\space\space\space \alpha_2 = 3$

Complimentary function CF $=e^{\alpha_1x}\Phi_1(y+m_1x)+e^{\alpha_2 x}\Phi_2(y+m_2 x)$

$=e^{0x}\Phi_1(y+x)+e^{3 x}\Phi_2(y-x)$

$=\Phi_1(y+x)+e^{3 x}\Phi_2(y-x)$

Now Particular Integrals corresponding to $xy$ and $7$ are as follows:

P.I.₁ $=\dfrac{1}{(D-D')(D+D'-3)}xy$

Taking $3D$ common from the denominator

$=\dfrac{1}{-3D\bigg(1-\dfrac{D'}{D}\bigg)\bigg(1-\dfrac{D}{3}-\dfrac{D'}{3}\bigg)}xy$

$=\dfrac{-1}{3D}\bigg(1-\dfrac{D'}{D}\bigg)^{-1}.\bigg(1-\dfrac{D}{3}-\dfrac{D'}{3}\bigg)^{-1}xy$

Using Binomial expansion (1+x)^-1 and solving we get

$=\dfrac{-1}{3}\bigg[\dfrac{1}{D}+\dfrac{1}{3}+\dfrac{2}{3}\dfrac{D}{D'}+\dfrac{1}{9}D+\dfrac{1}{3}D'+\dfrac{1}{9}DD'+\dfrac{D'}{D^2}+.......\bigg]xy$

On taking terms of D, D', DD' to the power 1 since in F(x, y), x and y are in power 1 only

P.I.₁$=-\dfrac{1}{3}\bigg[\dfrac{1}{2}x^2y+\dfrac{1}{3}xy+\dfrac{1}{3}x^2+\dfrac{1}{9}y+\dfrac{1}{3}x+\dfrac{1}{9}+\dfrac{1}{6}x^3\bigg]$

P.I.₂$=\dfrac{-1}{3D}\bigg(1-\dfrac{D'}{D}\bigg)^{-1}.\bigg(1-\dfrac{D}{3}-\dfrac{D'}{3}\bigg)^{-1}7$

$=\dfrac{-7}{3}\bigg[\dfrac{1}{D}+\dfrac{1}{3}+\dfrac{2}{3}\dfrac{D}{D'}+\dfrac{1}{9}D+\dfrac{1}{3}D'+\dfrac{1}{9}DD'+\dfrac{D'}{D^2}+.......\bigg]$

Taking terms whose numerator does not have D or D' as they will be equal to zero

$=\dfrac{-7}{3}\dfrac{1}{D}-\dfrac{7}{9}$

$=\dfrac{7x}{3}-\dfrac{7}{9}$

Complete Solution, $z=C.F+P.I._1+P.I._2$

$z=\Phi_1(y+x)+e^{3 x}\Phi_2(y-x)-\dfrac{1}{3}\bigg[\dfrac{1}{2}x^2y+\dfrac{1}{3}xy+\dfrac{1}{3}x^2+\dfrac{1}{9}y+\dfrac{1}{3}x+\dfrac{1}{9}+\dfrac{1}{6}x^3\bigg]+\dfrac{7x}{3}-\dfrac{7}{9}$