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.
$\endgroup$ 33 Answers
$\begingroup$$x^3 y^{′′′}−x^2 y^{′′}+2xy^′−2y=x^3$
Consider the substitution due to Euler:
$e^z = x$
You'll get
$y'=\dfrac{dy}{dx} =\dfrac{dy}{dz}\dfrac{dz}{dx}=e^{-z}\dfrac{dy}{dz}$
So $xy' = \dfrac{dy}{dz} $
Similarily you'll get
$x^2 y'' = \dfrac{d^2y}{dz^2}-\dfrac{dy}{dz}$ or $\mathcal{D}(\mathcal{D}-1)y$
And finally (as you might have guessed by now)
$x^3y''' =\mathcal{D}(\mathcal{D}-1)(\mathcal{D}-2)y$
Note that $\mathcal{D} = \dfrac{d}{dz}$ is our new operator.
Plugging this in gives
$\mathcal{D}(\mathcal{D}-1)(\mathcal{D}-2)y−\mathcal{D}(\mathcal{D}-1)y+2\mathcal{D}y−2y=e^{3z}$
Now factor $y$ and expand, then factor to get
$$\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$$
into a linear equation of constant coefficients.
$\endgroup$ 1 $\begingroup$I just wanted to try using Laplace : $$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?)
$\endgroup$ 2 $\begingroup$I can't help but feel like the answers given in this thread are way too complicated and although they work, the easier solution is just to look at the repeated root and realize that two of the linearly independent solutions are y1 = c1x and y2 = c2xln(x).
Why? It's easy to prove with reduction of order for a 2nd order linear homogeneous cauchy euler equation. I know your question is 4 years old, so I won't bother typing up a proof for nothing, but if anyone else stumbles upon this thread, you can message me and I'll explain in more detail.
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