An ordinary differential equation is said to be linear if $$F(t,y(t),...,y^{(n)}(t))=0$$ is linear in every derivative. I run into a little problem when using this definition for the equation $$y'y=0$$

because we have that $$F(t,\alpha y_1(t)+\beta y_2(t),y'(t))=(\alpha y_1(t)+\beta y_2(t))y'(t)=\alpha y_1(t)y't(t) + \beta y_2(t)y'(t)\\ =\alpha F(t,y_1(t),y'(t)) + \beta F(t,y_2(t),y'(t))$$

and vice versa for $y'(t)$. This shows that the ODE is linear. But an equivalent definition of linearity states that it has to have the form

$F(t,y(t),...,y^{(n)}(t))=\sum_{k=0}^n a_i(t)y^{(i)}(t) - g(t)=0$

With this definition the ODE is not linear anymore. Not sure what my mistake is there.


We have $F(t,y(t),y’(t))=yy’$. Note that $$F(\alpha t, \alpha y(t), \alpha y’(t)) =\alpha^2yy’ \neq \alpha yy’=\alpha F(t,y(t),y’(t))$$

and thus $F$ is not linear in its arguments, implying that it is not a linear differential equation using your first definition.


The problem is $y^{(n)}$ are dependent on $y$, so your definition is not sound. Derivative operators are linear by default, so the equation is linear if it is linear in $y$


$$ F(y) = y'y $$


$$ F(ay_1 + by_2) = (ay_1' + by_2')(ay_1+by_2) $$


$$ aF(y_1) + bF(y_2) = ay_1'y_1 + by_2'y_2 $$

which means

$$ F(ay_1+by_2) \ne aF(y_1) + bF(y_2) $$

Therefore $F(y)$ is not a linear operator.

  • $\begingroup$ Finally, thanks. Can you elaborate on what you mean by "not sound". Do you mean misleading, since all derivatives are dependend on the function itself? $\endgroup$ Mar 19 '19 at 17:05
  • 1
    $\begingroup$ By that, I mean the way that you applied it is not correct. $F(t,ay_1+by_2,y')$ doesn't make sense, since $y'$ cannot be independent of $y$. $\endgroup$
    – Dylan
    Mar 19 '19 at 17:15

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