Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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
add comment

2 Answers

up vote 3 down vote accepted

$x^3 y^{′′′}−x^2 y^{′′}+2xy^′−2y=x^3$

You should make the substitution due to Euler (all hail) - this is IMO the best way to solve this type of equations.

$e^z = x$

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.

share|improve this answer
    
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
add comment

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?)

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.