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I have equation of type: $$ a x^2 y(x)'' + bxy(x)' + cy = d(x) $$ it is called, in my text book, a second order Euler equation. They say that I need to put $e^t$ instead of $x$ and than $Y(x(t))=y(x)$

After this, they write that it is clear that: $$\begin{align} x\frac{dy}{dx}&=\frac{dY}{dt}\\ (x^2)y'' &= \frac{d^2Y}{dt^2}-\frac{dY}{dt} \end{align} $$ Please tell me what are they doing here.

thanks.

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can you tell us the name of the book and the page ? or can you provide the full text? –  user13838 Oct 8 '11 at 8:50
    
Also this has the full derivation. –  user13838 Oct 8 '11 at 8:51
    
the book is in Hebrew. –  Nahum Litvin Oct 8 '11 at 8:52
    
@percusse I found this link as well but I cannot understand how to use this in an exam. –  Nahum Litvin Oct 8 '11 at 8:53
    
in the link I provided, equation number (8) to (11), is the variable substitution $e^z=x$ or $z=\ln x$. You can use this in an exam if you know how to solve the resulting second order diff. eq. with constant coefficients. –  user13838 Oct 8 '11 at 9:14
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It's clear that $$\frac{dx}{dt} = x.$$

Using the chain rule on $Y(x(t))$, this implies that

$$\frac{dY}{dt} = \frac{dy}{dx}\frac{dx}{dt} = x \frac{dy}{dx}.$$

You now derive once more with respect to $t$:

$$\begin{align}\frac{d^2Y}{dt^2} &= \frac{dx}{dt} \frac{dy}{dx} + x \frac{d}{dt}\frac{dy(x(t))}{dx}.\\ &= x \frac{dy}{dx} + x \frac{d^2y}{dx^2} \frac{dx}{dt}\\ &= \frac{dY}{dt} + x^2 y^{''}(x). \end{align}$$

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I don't know what are they doing but I know that for Euler differential equation we can make substitution $x=e^t$, and after that we can write following:

$x'_t=(e^t)'_t \Rightarrow \frac{dx}{dt}=e^t \Rightarrow dx=e^tdt$

Since we know that $y'_x=\frac{dy}{dx}$ ,we can write:

$y'_x=\frac{dy}{e^tdt}=-e^t\frac{dy}{dt}=-e^ty'_t$

Similarly we can show that:

$y''_x=(y''_t-y'_t)e^{-2t}$

Now if you substitute these expressions into ODE you will get:

$ay''_t+(b-a)y'_t+cy=d(t)$, which can be solved if we first solve homogeneous equation:

$ay''_t+(b-a)y'_t+cy=0$

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