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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?2012-07-08
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    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.2012-07-08
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    So $y'=y$ isn't linear, even though it's the same as $y'-y=0$, which is linear?2012-07-09
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    My question was whether $p_0(t)$ could be in front of $y^n$2012-07-09
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    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?2012-07-10
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    Well, $p_i(t) = 0$.2012-07-10
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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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