# Multiplying out a quadratic equation of differential factors acting on a function

I am having trouble understanding a multiplying out of a quadratic equation with differential factors:

from the following video: $$(D + A(x))(D + B(x))y(x) = 0\\ (D^2 + AD + AB + B^{'} + BD)y = 0 \\$$ What I don't get is his remark on the last multiplication two terms. It looks like he is using the multiplication rule for differentiation, but in the end, the multiplied out parenthesis is again acting on y, so for me it looks like a double use.

I see that $$B^{'} = DB$$ but I still do not get why two terms are produced?

• Note that $BDy \ne DBy$ the second is the derivative of a product and gives two terms. The first term is a single one. – Aryadeva Jun 30 at 17:50

$$(D + A(x))(D+B(x))y(x)=0$$ (where D is the differentiation operator $$(DX(x))=dX/dx$$)

$$\implies(D + A(x))(Dy(x)+B(x)y(x))=0$$

If you now expand you get $$D(D(y(x)))+D(B(x)y(x))+D(y(x))*A(x)+A(x)B(x)y(x)$$

The term $$D(B(x)y(x))$$ can be written as :

$$D(B(x)y(x))$$=$$D(B(x))*y(x)+D(y(x))*B(x)$$ (by the product rule)

So if you plug that in the previous equation you get

$$D(D(y(x)))+D(B(x))*y(x)+D(y(x))*B(x)+D(y(x))*A(x)+A(x)B(x)y(x)$$

And now if you pull $$y$$ out you get

($$D^2+B'+B(x)+A(x)D + A(x)B(x))y(x)$$

• Thank you. Obviously, my dyslexia has struck again as I had trouble to see $D(By) = DBy + BDy = (DB + BD)y = (B^{'} + BD)$. Even by writing this right now, it's hard for me to see the difference between DB and BD. – Johannes Maria Frank Jun 30 at 17:45

Omitting the argument $$(x)$$, we have $$(D+b)y=Dy+by$$

and

$$(D+a)(D+b)y=(D+a)(Dy+by)=D^2y+D(by)+aDy+aby \\=D^2y+bDy+b'y+aDy+aby \\=(D^2+(a+b)D+(b'+ab))y.$$

$$(D + A(x))(D + B(x))y(x) = 0$$ $$(D + A(x))(y' + B(x)y(x)) = 0$$ $$D(y' + B(x)y(x)) + A(x)(y' + B(x)y(x)) = 0$$ $$y'' + D( B(x)y(x)) + A(x)y' + A(x)B(x)y(x)) = 0$$ You differentiate a product you have a sum of two terms. $$y'' + B'(x)y(x)+B(x)y'(x) + A(x)y' + A(x)B(x)y(x) = 0$$ $$y'' +(B(x) + A(x))y' + (B'(x)+A(x)B(x)y(x) = 0$$