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As I understand it, the differential equation

$$y^{(n)} + p_1(t)y^{(n-1)} + ... + p_{(n-1)}(t)y^{(1)} = f$$

is linear because the left hand side can be written as $L[y]$ where $L$ is a linear operator. So why isn't

$$p_0(t)y^{(n)} + p_1(t)y^{(n-1)} + ... + p_{n-1}(t)y^{(1)} = f$$

considered linear?

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Who says it isn't? – Gerry Myerson Jul 8 '12 at 6:11
Well my textbook said that a linear differential equation is of the form $y^{(n)} + p_1(t)y^{(n-1)} + ... + p_{(n-1)}(t)y^{(1)} = f$ which led me to believe that only differential equations like that were linear. – Dom Jul 8 '12 at 23:26
So $y'=y$ isn't linear, even though it's the same as $y'-y=0$, which is linear? – Gerry Myerson Jul 9 '12 at 0:15
My question was whether $p_0(t)$ could be in front of $y^n$ – Dom Jul 9 '12 at 19:38
Your question was whether it had to look exactly like what's in the book for it to be linear. $y'=y$ doesn't look exactly like what's in the book. Does that mean it isn't linear? – Gerry Myerson Jul 10 '12 at 1:38

The second equation is linear as well. In general, a differential equation is linear, if $y_1(t)$ and $y_2(t)$ both satisfy the homogeneous differential equation, then so does $\alpha y_1(t) + \beta y_2(t)$, where $\alpha, \beta \in \mathbb{R}$ i.e. if $L(y_1(t)) = 0$ and $L(y_2(t)) = 0$, then we also have that $$L(\alpha y_1(t) + \beta y_2(t)) = 0$$

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