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.