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I'm just wondering, I've come across a Second Order DE Question which says:

Find the general solution of the ODE: $ 2 \frac {d^2y}{dx^2} + 5 \frac {dy}{dx} − 3y = 0. $

I'm just wondering how this differs from the notes I've seen which use $ y''$ I'm thinking that it's the same but I just want to make sure

Thanks

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It does not differ at all, it's just a different notation.

It's the same as : $2y'' + 5y' - 3y = 0$

Any derivative of a function, let's say $y(x)$ with respect to $x$, can be written as $y^{(n)} = \frac{d^ny}{dx^n}$. Note that until the third derivative, you use the $'$ notation : $y^{(1)} = y',y^{(2)} = y'', y^{(3)} = y''' $ and for higher derivatives, you use the $y^{(n)} $ notation, where n is the grade of the derivative.

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Both of them are notations to show the order of derivative.

You can replace $\frac{d^2y}{dx^2}$ with $y''$. Similarly $\frac{dy}{dx}$ with $y'$.

So your equation becomes,

$2y''+5y'-3y=0$

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