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The L1 Mat-1.1010 -course here has taught me the linearity conditions $f(a x)=a f(x)$ and $f(a+b)=f(a)+f(b)$. I want to generalize it, some quite irrelevant slow investigation here.

It requires time to verify statements like below, source here.

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According to Wikipedia here, $Ly=f$ and

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but I cannot see this kind of statements fast, unless going through the conditions one-by-one. So are there fast ways to check whether some differential is linear or not?

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    It's handy to remember that sums of linear operators are linear, so you can break stuff down.2012-09-20

1 Answers 1

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A linear differential equation (of the second order) is of the form

$a(x)\frac{d^2y}{dx^2}+b(x)\frac{dy}{dx}+c(x)y=d(x).$

$y$ and its derivatives may not appear with a power or as the argument of a function.

Only $1.4$ fulfills this (with $a(x)=x^2,b(x)=0,c(x)=2,d(x)=2$).

The rule generalizes to all orders.


As you can check, with $z=\lambda y$ where $y$ is a solution,

$a(x)\frac{d^2z}{dx^2}+b(x)\frac{dz}{dx}+c(x)z=\lambda a(x)\frac{d^2y}{dx^2}+\lambda b(x)\frac{dy}{dx}+\lambda c(x)y=\lambda d(x),$

and with $z=y_1+y_2$ where $y_1,y_2$ are two solutions,

$a(x)\frac{d^2z}{dx^2}+b(x)\frac{dz}{dx}+c(x)z\\ =a(x)\frac{d^2y_1}{dx^2}+b(x)\frac{dy_1}{dx}+c(x)y_1+a(x)\frac{d^2y_2}{dx^2}+b(x)\frac{dy_2}{dx}+c(x)y_2=2d(x),$

so that in both cases $z$ is also a solution (provided you account for the factors in the RHS).


$1.5$ isn't linear because $z^2=(\lambda y)^2=\lambda^2y^2\ne\lambda y^2$.

$1.6$ isn't linear because $\sin z=\sin\lambda y\ne\lambda\sin y$.