Transformation of a Cauchy-Euler equation

[jawhnay]

Can anyone explain to me how I would go about transforming a Cauchy-Euler equation for an equation such as:
x²y'' - xy' = ln x

I know you have to start with x = e^t or t = ln x however I'm not sure what to do next...

 

[Gengar]

Well making the substitution [tex]x=e^t[/tex] does indeed get the solution, you should get via chain rule:

[tex]\frac{d}{dx}=e^{-t}\frac{d}{dt}, \frac{d^2}{dx^2}=e^{-2t}(\frac{d^2}{dt^2}-\frac{d}{dt})[/tex]

which when substituted into your original equation should yield an inhomogeneous 2nd order ODE with constant coefficients which we can do via the standard method of finding complementary solutions and then a particular solution.

 

[JJacquelin]

With z(x)=y'(x), the ODE is of the first order :
x²z'+xz=ln(x)
Solving x²z'-xz=0 leads to z=c*x where c=constant
Then bing back z=u(x)*x into x²z'-xz=ln(x)
which leads to u' =ln(x)/(x^3)
u(x)=-(2*ln(x)+1)/(4*x²)+C where C=constant
y'=z(x)=u(x)*x=-(2ln(x)+1)/(4*x)+C*x
y= -(ln(x)+1)*ln(x)/4 +A*x² +B where A, B are constants.

 

[Mark44]

Thread closed. See the duplicate post at https://www.physicsforums.com/showthread.php?t=680980.

 

[blue_raver22]

To transform a Cauchy-Euler equation, you need to make a substitution that will turn the equation into a standard form that can be easily solved. In this case, since the equation involves a natural logarithm, it would be helpful to substitute x = e^t. This will transform the equation into one that involves only t and its derivatives.

So, let's start by making the substitution x = e^t. This means that dx/dt = e^t and d^2x/dt^2 = e^t. Substituting these into the original equation, we get:

(e^t)^2y'' - e^t y' = ln(e^t)

Simplifying, we get:

e^2t y'' - e^t y' = t

Now, we can divide both sides by e^t to get the equation in standard form:

y'' - y' = t/e^t

This is now a standard form for a Cauchy-Euler equation, which can be solved using various methods such as the method of undetermined coefficients or the method of variation of parameters.

I hope this helps to explain the process of transforming a Cauchy-Euler equation. Keep in mind that the specific substitution you make may vary depending on the form of the equation, so it's important to carefully analyze the equation and choose a substitution that will simplify it as much as possible.