# Thread: transform the Cauchy-Euler equation

1. ## transform the Cauchy-Euler equation

ok in this problem im supposed to use the substitution x = e^t to transform the given Cauchy-Euler equation to a differential equation with constant coefficients.

x^2y'' +9xy' -20y=0

im not sure how to start this problem

maybe dy/dx= (dy/dt)(dt/dx) = (dy/dt)e^t

so then could i say d2y/dx^2 = (e^t)d2y/dt^2<---- ?

maybe then i could substitute back into the equation? i dunnno..

any help would be appreciated thanks in advance.

2. $\displaystyle x=e^t\Rightarrow t=\log x \Rightarrow y'(x)=x'(t)\cdot{(1/x)=x'(t)e^{-t}}$

Then,

$\displaystyle y''(x)=\ldots=(y''(t)-y'(t))e^{-2t}$

P.D. Another method to solve the equation is to try solutions of the form $\displaystyle y=x^k$ .

3. ok so i saw my mistake in taking the second derivative and this is what i have so far

$x^{2}(e^{t}\frac{\mathrm{d} y}{\mathrm{d} t}+\frac{\mathrm{d}^2y }{\mathrm{d} t^2}e^{t}) + 9x(e^{t}\frac{\mathrm{d} y}{\mathrm{d} t})-20y=0$

am i on the right track?

4. this problem is supposed to reduce down to m^2 + 8m -20 = 0???

5. For Cauchy-Euler equations, I've always preferred FernandoRevilla's PD in post # 2. You get the characteristic equation immediately by plugging that ansatz into the DE.

6. The problem with simply taking $\displaystyle y= x^r$, as both Fernando Revilla and Ackbeet suggest, is that it not clear what to do with "special cases". For example, consider the differential equation $\displaystyle x^2y''- xy'+ y= x+ ln(x)$. Taking $\displaystyle y= x^r$ the equation reduces to $\displaystyle (r^2- r)x^r- rx^r+ x^r= (r^2- 2r+ 1)x^r= 0$ which gives the characteristic equation $\displaystyle r^2- 2r+ 1= (r- 1)^2= 0$. Okay, one solution is $\displaystyle y= x^{1}$ but what is the other, independent, solution we need? And what should we do about that right side, x+ ln(x)?

If we know that the substitution $\displaystyle x= e^t$ changes the Euler-Cauchy equation to the constant-coefficients equation with the same characteristic equation, we know that the solution to the assosciated homogeneous equation, in terms of t, is $\displaystyle Ce^{t}+ Dte^{t}$, which tells us that the corresponding solution to the original equation is $\displaystyle Cx+ Dxln(x)$. Also, the right hand side, in terms of t, is $\displaystyle e^t+ t$, which, by "undetermined coefficients" has particular solution $\displaystyle \frac{1}{2}t^2e^{t}+ t+ 2$ so that the entire general solution to the original equation is
$\displaystyle y(x)= Cx+ Dxln(x)+ \frac{1}{2}(ln(x))^2x+ ln(x)+ 2$.

Of course, we could learn variations of all those rules for Cauchy-Euler equations but it is much simpler to know how to convert such an equation to a constant-coefficients equation.

7. Originally Posted by slapmaxwell1
x^2y'' +9xy' -20y=0
My notation is probably going to make several members cringe, but this is the way I learned it...

As was suggested let
$\displaystyle \displaystyle x = e^t \implies dx = e^t~dt \implies \frac{d}{dx} = e^{-t} \frac{d}{dt}$

So looking at your original equation as
$\displaystyle \displaystyle x^2 \frac{d}{dx} \left ( \frac{d}{dx} \right ) y + 9x \frac{d}{dx}y - 20y = 0$

gives us
$\displaystyle \displaystyle e^{2t} \cdot e^{-t} \frac{d}{dt} \left ( e^{-t} \frac{d}{dt} \right ) y + 9e^t \cdot e^{-t} \frac{d}{dt}y - 20y = 0$

So,after simplifying:
$\displaystyle \displaystyle e^{2t} \cdot \left ( -e^{-2t} \frac{dy}{dt} + e^{-2t} \frac{d^2y}{dt^2} \right ) + 9 e^t \cdot e^{-t} \frac{dy}{dt} - 20y = 0$

And finally:
$\displaystyle \displaystyle \frac{d^2y}{dt^2} + 8 \frac{dy}{dt} - 20y = 0$

and you can take over from here.

-Dan

8. ok im still a little confused, i dont see how this is true?

$\frac{\mathrm{d}^2 y}{\mathrm{d} x^2}=\frac{1}{x^2}(\frac{\mathrm{d}^2 y}{\mathrm{d} t^2}-\frac{\mathrm{d} y}{\mathrm{d} t})$

i get how to find dy/dx, but not the second derivative...

$\frac{\mathrm{d}^2 y}{\mathrm{d} x^2}=\frac{-1}{x^2}\frac{\mathrm{d} y}{\mathrm{d} t}+\frac{1}{x}\frac{\mathrm{d}^2 y}{\mathrm{d} t^2}$

i keep getting a 1/x not 1/x^2...but im just using the chain rule

9. Originally Posted by slapmaxwell1
ok im still a little confused, i dont see how this is true?

$\frac{\mathrm{d}^2 y}{\mathrm{d} x^2}=\frac{1}{x^2}(\frac{\mathrm{d}^2 y}{\mathrm{d} t^2}-\frac{\mathrm{d} y}{\mathrm{d} t})$

i get how to find dy/dx, but not the second derivative...

$\frac{\mathrm{d}^2 y}{\mathrm{d} x^2}=\frac{-1}{x^2}\frac{\mathrm{d} y}{\mathrm{d} t}+\frac{1}{x}\frac{\mathrm{d}^2 y}{\mathrm{d} t^2}$

i keep getting a 1/x not 1/x^2...but im just using the chain rule
One problem stems from leaving that x there.

Example:
$\displaystyle x = e^t$

so
$\displaystyle \displaystyle \frac{d^2y}{dx^2} = x^2 \left ( \frac{d^2y}{dt^2} - \frac{dy}{dt} \right )$

should be
$\displaystyle \displaystyle \frac{d^2y}{dx^2} = e^{-2t} \left ( \frac{d^2y}{dt^2} - \frac{dy}{dt} \right )$

If the problem is that you need a full derivation
$\displaystyle \displaystyle \frac{d}{dx} = e^{-t} \frac{d}{dt}$

So the second derivative of y with respect to x becomes:
$\displaystyle \displaystyle \frac{d}{dx} \left ( \frac{dy}{dx} \right ) = \left [ e^{-t} \frac{d}{dt} \right ] \left [ e^{-t} \frac{d}{dt} \right ]y = \left [ e^{-t} \frac{d}{dt} \right ] \left [ e^{-t} \frac{dy}{dt} \right ]$

Now use the product rule
$\displaystyle \displaystyle = e^{-t} \left [ -e^{-t} \frac{dy}{dt} + e^{-t} \frac{d^2y}{dt^2} \right ]$

-Dan

10. ok i see it, i was taking the derivative wrong. i went and reviewed implicit differentiation....thanks you guys!

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# substitution for e^x in caushy's differential equation

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