Answer to Question #218632 in Differential Equations for rahul

Question

Particular integral of (D^2 +3DD' + 2D'^2) = 12xy

Answer 1

$(𝐷^ 2 +3DD'+ 2𝐷'^2)𝑧 = 12𝑥𝑦.$

$𝐷^ 2 +3DD'+2 𝐷'^2=(D+D')(D+2D')$

$(𝐷^ 2 +3DD'+ 2𝐷'^2)𝑧 =(D+D')(D+2D')z$

$(D+D')(D+2D')z=12xy$

The auxiliary equation of the given equation is

$(k+1)(k+2)=0$

$k_1=-1, k_2=-2$

Than the complementary function of the given equation is

$u=\varphi_1(y-x)+\varphi_2(y-2x)$

Than partial integral

$P.I.=\dfrac{1}{(D+D')(D+2D')}(12xy)$

$=\dfrac{1}{D(1+\dfrac{D'}{D})D(1+\dfrac{2D'}{D})}(12xy)$

$=\dfrac{1}{D^2}\bigg[1+\dfrac{D'}{D}\bigg]^{-1}\bigg[1+\dfrac{2D'}{D}\bigg]^{-1}(12xy)$

$\dfrac{1}{1+t}=1-t+t^2-t^3+t^4-t^5+..., |t|<1$

$P.I.=\dfrac{1}{D^2}\bigg(1-\dfrac{D'}{D}+...\bigg)\bigg(1-\dfrac{2D'}{D}+...\bigg)(12xy)$

$=\dfrac{1}{D^2}\bigg(1-\dfrac{3D'}{D}+\dfrac{2D'^2}{D^2}+...\bigg)(12xy)$

$=\dfrac{1}{D^2}\bigg(12xy-\dfrac{3}{D}\dfrac{\partial}{\partial y}(12xy)$

$+\dfrac{2}{D^2}\dfrac{\partial^2}{\partial y^2}(12xy)+...\bigg)$

$=\dfrac{1}{D^2}\bigg(12xy-\dfrac{36}{D}x+\dfrac{2}{D^2}(0)+...\bigg)$

$=\dfrac{1}{D^2}\bigg(12xy-36\int xdx\bigg)$

$=\dfrac{1}{D}\bigg(\int(12xy-18x^2)dx\bigg)$

$=\dfrac{1}{D}(6x^2y-6x^3)=\int(6x^2y-6x^3)dx$

$=2x^3y-\dfrac{3}{2}x^4$

$P.I.=2x^3y-\dfrac{3}{2}x^4$

$z=\varphi_1(y-x)+\varphi_2(y-2x)+2x^3y-\dfrac{3}{2}x^4$

where $\varphi_1$ and $\varphi_2$ are arbitrary functions.