I have to solve the problem: $(x^2-x+1)\frac{d^2y}{dx^2}-(x^2+x)\frac{dy}{dx}+(x+1)y=0$.

I've tried using the substitution $y=x^r$, and that gives me a long string of r's and x's and exponents that I'm not sure what to do with: $(r-1)x^{r-2}(r(x^2-x+1)-x^2(x+1))=0$. From what I understand of the Cauchy-Euler method, I'm supposed to find the characteristic equation, but I don't see how to from the initial result.

  • $\begingroup$ This is actually not a Cauchy-Euler Equation $\endgroup$
    – Dylan
    Mar 30, 2015 at 19:25
  • $\begingroup$ Oh, why is that? $\endgroup$
    – camdroid
    Mar 31, 2015 at 1:43
  • $\begingroup$ Check the definition. A Cauchy-Euler equation has the form $$ax^2\frac{d^2y}{dx^2} + bx\frac{dy}{dx} + cy = 0$$ $\endgroup$
    – Dylan
    Mar 31, 2015 at 17:58
  • $\begingroup$ All right, that makes sense. Is there another term for this type of equation? $\endgroup$
    – camdroid
    Apr 1, 2015 at 0:35
  • $\begingroup$ What do you mean by "another term"? $\endgroup$
    – Dylan
    Apr 1, 2015 at 3:30

1 Answer 1


An obvious solution is $y=x$

We use the classical method to reduce the degree of the ODE which consists in the change of function $y(x)=xf(x)$

$y'=xf'+f$ and $y''=xf''+2f'$

$$(x^2-x+1)(xf''+2f')-(x^2+x)(xf'+f)+(x+1)xf=0$$ $$(x^2-x+1)xf''+(-x^3+x^2-2x+2)f'=0$$ Let $g(x)=f'$ $$(x^2-x+1)xg'+(-x^3+x^2-2x+2)g=0$$ This method reduces the ODE to a separable first order one. $$\frac{g'}{g}=\frac{x^3-x^2+2x-2}{(x^2-x+1)x}$$ $$\ln(g)=x-2\ln(x)+\ln(x^2-x+1)+constant$$ $$g=c_1e^x\frac{x^2-x+1}{x^2}$$ $$f=c_1e^x\left(1-\frac{1}{x}\right)+c_2$$ $$y=c_1e^x(x-1)+c_2x$$

  • $\begingroup$ Thanks for the detailed explanation, it was very helpful! $\endgroup$
    – camdroid
    Mar 30, 2015 at 19:07
  • $\begingroup$ How is $y = x$ an "obvious" solution? What's the basis for guessing it? $\endgroup$
    – Dylan
    Mar 30, 2015 at 19:24
  • $\begingroup$ It takes only a few minutes to check the most simple usual functions and see if, by chance, one of them is a solution. Of course, if not, we have to try more complicated methods. $\endgroup$
    – JJacquelin
    Mar 30, 2015 at 20:11

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