# Third order Cauchy-Euler differential equation

I need to solve this: $$x^3y''' - x^2y'' + 2xy' - 2y = x^3$$

I know that first I have to solve: $$E = x^3y''' - x^2y'' + 2xy' - 2y = 0$$

I choose $y = x^r$. By feeding that to $E$ it will lead me to the following characteristic equation: $$(r-2)(r-1)^2=0$$

Now if all roots would be distinct the solution would be simple, but with repeating roots how do I approach this equation?

I know that the correct answer must be: $$y(x) = c_3 x^2+c_1 x+c_2 x ln(x)+x^3/4$$

I don't know how to get to the $c_2xln(x)$ part. I know that I have to use Wronskian matrix to get the $x^3/4$ part.

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Go to wolframalpha.com/input/?i=x^3*y%27%27%27-x^2*y%27%27%2B2*x*y%27-2*y%3‌​Dx^3 you can find the solution and all the steps. – Riccardo.Alestra Feb 15 '12 at 8:02
I did that and WA says this: 'Because of repeated root r = 1 we have the solutions y1 = c1x and y2 = c2xln(x)'. Now my question is Why?. :-s – Dr.Optix Feb 15 '12 at 15:41
@Dr.Optix Read my answer, you'll see why. – Pedro Tamaroff Feb 15 '12 at 22:15

$x^3 y^{′′′}−x^2 y^{′′}+2xy^′−2y=x^3$

Consider the substitution due to Euler:

$e^z = x$

You'll get

$y'=\dfrac{dy}{dx} =\dfrac{dy}{dz}\dfrac{dz}{dx}=e^{-z}\dfrac{dy}{dz}$

So $xy' = \dfrac{dy}{dz}$

Similarily you'll get

$x^2 y'' = \dfrac{d^2y}{dz^2}-\dfrac{dy}{dz}$ or $\mathcal{D}(\mathcal{D}-1)y$

And finally (as you might have guessed by now)

$x^3y''' =\mathcal{D}(\mathcal{D}-1)(\mathcal{D}-2)y$

Note that $\mathcal{D} = \dfrac{d}{dz}$ is our new operator.

Plugging this in gives

$\mathcal{D}(\mathcal{D}-1)(\mathcal{D}-2)y−\mathcal{D}(\mathcal{D}-1)y+2\mathcal{D}y−2y=e^{3z}$

Now factor $y$ and expand, then factor to get

\eqalign{ & \left( {{\mathcal{D}^3} - 4{\mathcal{D}^2} + 5\mathcal{D} - 2} \right)y = {e^{3z}} \cr & {\left( {\mathcal{D} - 1} \right)^2}\left( {\mathcal{D} - 2} \right)y = {e^{3z}} \cr}

So now solve the homogeneous equation:

$${\left( {\mathcal{D} - 1} \right)^2}\left( {\mathcal{D} - 2} \right)y = 0$$

This gives the complementary solution in terms of $z$ or $\log x$

$${y_c} = {c_1}z{e^z} + {c_2}{e^z} + {c_3}{e^{2z}}$$

$${y_c} = {c_1}x\log x + {c_2}x + {c_3}{x^2}$$

You can easily get the particular with the original equation by assuming a solution $y=Ax^3$ and finding $A$.

ADD: In general, the substitution $x = e^z$ will transform

$$\sum\limits_{k = 0}^n {{a_k}{x^k}{D^k}} y = F$$

into a linear equation of constant coefficients.

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Sorry to bump this up but why did you assume $y=Ax^3$ instead of $y=Ax^3+Bx^2+Cx+D$? Thank you! – drawar Sep 27 '13 at 8:44

I just wanted to try using Laplace : $$x^3y''' - x^2y'' + 2xy' - 2y = x^3$$ $$\implies y''' - y'' + 2y' - 2y = e^{-3t}$$ See above link for substitution details :: $x = e^{t}$ Henceforth, y prime denotes differentiation wrt t $$\implies L(y''') - L(y'') + 2L(y') - 2L(y) = L(e^{-3t})$$ Let $L(y) = \phi(s)$

$$\implies (s^3\phi(s)-s^2f(0) -sf'(0) - f''(0)) - (s^2\phi(s)-sf(0) - f'(0)) + 2(s.\phi(s) - f(0)) - 2\phi(s) = \frac{1}{s+3}$$

$$\phi(s)(s^3 - s^2 + 2s - 2) - f(0)(s^2 -s+2) -f'(0)(s-1) - f''(0) = \frac{1}{s+3}$$ $$\phi(s)(s^3 - s^2 + 2s - 2) = f(0)(s^2 -s+2) +f'(0)(s-1) + f''(0) + \frac{1}{s+3}$$ $$\phi(s)= \frac{f(0)(s^2 -s+2) +f'(0)(s-1) + f''(0) + \frac{1}{s+3}}{(s^3 - s^2 + 2s - 2) }$$

$$\phi(s)= \frac{f(0)(s^2 -s+2)}{(s^3 - s^2 + 2s - 2)} +\frac{f'(0)(s-1)}{(s^3 - s^2 + 2s - 2)} + \frac{f''(0)}{(s^3 - s^2 + 2s - 2) } +\frac{1}{(s+3)(s^3 - s^2 + 2s - 2) }$$ I hope I haven't made any errors. This should give you an answer. I was going to throw it away but posted it hoping someone might find it useful (or would they?)

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If you could factor out the denominators I guess I'd work, but still, the LT of $x \log x$ involves $\gamma$ so it's kind of a problem and long work. – Pedro Tamaroff Feb 18 '12 at 23:07
@Peter, I factored it in MATLAB and Yes, It is a major pain. As I said, it was an attempt to do it using Laplace. – Inquest Feb 19 '12 at 6:44