# Definition linear ODE

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 $$y$$

Let

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

Then

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

while

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

which means

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

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

• 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? Mar 19 '19 at 17:05
• 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 $y$. Mar 19 '19 at 17:15