Transformation of a Cauchy-Euler equation

  • Thread starter jawhnay
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  • #1
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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 = et or t = ln x however I'm not sure what to do next...
 

Answers and Replies

  • #2
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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.
 
  • #3
798
34
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.
 

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