I am told that linear differential equations are those in which neither the function nor its derivatives occur in products, powers or nonlinear functions.

The following function obviously occurs in a product with $t^2$ but it is classified as a linear differential equation. How is this so? Is my reasoning incorrect? If so, why?

$\dfrac{\partial^4F}{\partial x \partial y^3} = t^2F$

Thank you.

up vote 1 down vote accepted

In a linear differential equation, the function it self can be multiplied by any (function, non-linear or not) of the variables. For example, the following equation

$$ (8x^2)y''(x) + \sin(x)y'(x) + e^x y(x) = 0 $$

is a perfectly legitimate linear ODE for $y$ even though $y(x)$ is multiplied by $e^x$. Note that if $y_1,y_2$ are solutions of the ODE then so is $a_1 y_1 + a_2 y_2$ which is what you want from a linear (homogeneous) equation.

In your case, I assume that $t$ is a variable and $F = F(x,y,t)$ and so your equation is also a linear equation for $F$ and satisfies the property that if $F_1,F_2$ are solutions of the equation then so are $a_1 F_1 + a_2 F_2$.

  • 1
    As a counterpoint, $\frac{\partial^4 F}{\partial x\partial y^3} = F^2$ is not linear, for example. – arkeet Oct 11 '16 at 2:23
  • @arkeet Because we are obviously multiplying F by itself? – The Pointer Oct 11 '16 at 2:44
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    @ThePointer Yep. The equation $\frac{\partial^4 F}{\partial x \partial y^3} = \sin(F)$ wouldn't be linear either. The equation $\sin \left( \frac{\partial^4 F}{\partial x \partial y^3} \right) = F$ wouldn't be linear either because a non-linear function ($\sin$) is applied to one of the derivatives. – levap Oct 11 '16 at 2:51
  • @levap This makes sense. Thank you! – The Pointer Oct 11 '16 at 2:58

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