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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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    If $p_i(t)$ is identically zero, then you have $y^{(n)}=f$ where presumably $f$ is a function of $t$, not of $y$. So you don't get $y'=y$ by letting some coefficient(s) vanish.2012-07-10

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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 $L(\alpha y_1(t) + \beta y_2(t)) = 0$