Why is this a linear differential equation?

[SMA_01]

Why is y'-2xy=x a linear differential equation? I thought it would be nonlinear due to the 2xy...?

 

[Ray Vickson]

Why is y'-2xy=x a linear differential equation? I thought it would be nonlinear due to the 2xy...?

It is linear *in y*; it would be a nonlinear DE if it contained things like y^2, exp(y), 1/(1+y), etc. Another way to see it is: if y1 and y2 are two solutions and a, b are constants, then the linear combination a*y1 + b*y2 is also a solution. That would generally fail for a nonlinear DE.

RGV

 

[Ray Vickson]

It is linear *in y*; it would be a nonlinear DE if it contained things like y^2, exp(y), 1/(1+y), etc. Another way to see it is: if y1 and y2 are two solutions and a, b are constants, then the linear combination a*y1 + b*y2 is also a solution. That would generally fail for a nonlinear DE.

RGV

Sorry: I mean that the above linear combination property is true of the _homogeneous_ equation, with 0 on the right-hand-side; it may, of course, fail for a nonzero right-hand-side. Also: I should have mentioned that a DE is nonlinear as well if it has terms nonlinear in y and/or y', such as y*y' or (y')^2, etc.

RGV

 

[A. Bahat]

Another way of looking at it is to consider the operator $L=\frac{d}{dx}-2x$, so that the differential equation becomes $Ly=x$. Then we say the differential equation is linear if that operator $L$ is linear, i.e. $$L(f+g)=L(f)+L(g)$$ and $$L(cf)=cL(f)$$ for all (suitably smooth) functions $f$ and $g$ and constants $c$ (where addition of functions and multiplication of a function by a constant are defined point-wise as usual). This is equivalent to what Ray Vickson just said: if you consider the equation $Ly=0$, then (as a result of the linearity of $L$) any linear combination of solutions is also a solution.