Answer to Question #213751 in Differential Equations for Rider

Question

(D²-3DD'+2D'²)z=(2+4x)e^x+2y.

Answer 1

$(D^{2}-3DD^{'}+2D^{'})z=(2+4x)e^{x+2y}$

Now , to solve this type of differential equation , we let the following terms of equation as -

$D=m\ D^{'}=1$

Now the given Differential equation ,will be form as -

So Auxiliary equation will become as -

$=m^{2}-3m+2=0$

$=m^{2}-2m-m+2=0$

$=m(m-2)-1(m-2)=0$

$=(m-1)(m-2)=0$

So this is case where roots are different -

So CF can be written as -

$CF=f_{1}(y+x)+f_{2}(y+2x)$

Now PI of the Differential equation can be given as -

$PI=\dfrac{1}{(D^{2}-3DD^{'}+2D^{'})}(2+4x)e^{x+2y}$

$PI=2\dfrac{1}{(D^{2}-3DD^{'}+2D^{'})}e^{x+2y}+4\dfrac{1}{(D^{2}-3DD^{'}+2D^{'})}xe^{x+2y}$

$PI=2\dfrac{1}{(D^{2}-3DD^{'}+2D^{'})}e^{x+2y}$

Now , this is case of $\phi(ax+by)$ $;F(a,b)\ {\neq}0$

$PI=2\dfrac{1}{(1^{2}-3(1)(2)+2(2)}\iint e^{u}dudu$

$PI=-2e^{x+2y}$

Now PI of second function , we get -

$PI=4\dfrac{1}{(D^{2}-3DD^{'}+2D^{'})}xe^{x+2y}$

Now , this is a case of $F(x,y)=e^{ax+by},V,$ where V is a function of $x$ and y .

$PI=\dfrac{1}{\phi(D,D^{'})}e^{ax+by}.V=e^{ax+by}\dfrac{1}{\phi(D+a,D^{'}+b)}V$

$=e^{x+2y}$ $\dfrac{1}{((D+1)^{2}-3(D+1)(D^{'}+2)+2(D^{'}+2)}$ $x$

$=e^{x+2y}\dfrac{1}{D^{2}-D^{'}-4D-1}x$

$=e^{x+2y}\dfrac{1}{D^{2}}[1-(\dfrac{D^{'}}{D^{2}}+\dfrac{4}{D}+\dfrac{1}{D^{2}})]^{-1}x$ $=\dfrac {e^{x+2y}x^{3}}{6}$

Now complete solution is given as -

$z=CF+PI$ $=f_{1}(y+x)+f_{2}(y+2x)+\dfrac {e^{x+2y}x^{3}}{6}$