D.e inhomogneous non constant coeff

**i_zz_y_ill** · May 18th 2009, 04:56 AM

The question is find the solution of:

$x^2\frac{dy}{dx}-x\frac{dy}{dx}-8y=1+x$

clearly the coefficients in what i would write down to be the auxilliary equation are not constants. Ive been told by my solution to use y= $x^m$ but i dont see how this gives(in the solutions) $m(m-1)-m-8=x^2-2m-8=(x-4)(x+2)0$
giving m=4 or m=-2. and the corrosponding complementary function. Does anyone have any ideas or could show me how the method actually cancels out the non constant coefficeints? thanks. oh i think you are supposed to use liebnitz?? not sure

**Spec** · May 18th 2009, 05:53 AM

Cauchy-Euler equation - Wikipedia, the free encyclopedia

One way to solve it is to set $t=\ln x,\ x > 0$ (if $x < 0$ use $t=\ln -x$ instead)

Let $y(x)=z(t(x))$

Then $\frac{dy}{dx}=\frac{dz}{dt}\frac{dt}{dx}=\frac{1}{ x}\frac{dz}{dt}$ and $xy'(x)=z'(t)$

$\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{1}{x}\fr ac{dz}{dt} \right)=-\frac{1}{x^2}\frac{dz}{dt}+\frac{1}{x}\frac{d}{dx} \left( \frac{dz}{dt}\right)=-\frac{1}{x^2}\frac{dz}{dt}+\frac{1}{x^2}\frac{d^2z }{dt^2}$

We have that: $x^2y''(x)=z''(t)-z'(t)$

So instead you need to solve $z''(t)-2z'(t)-8z(t)=1+e^t$

**chisigma** · May 18th 2009, 06:38 AM

The complete equation is...

$x^{2}\cdot y^{''} - x \cdot y^{'} - 8 \cdot y = 1 + x$ (1)

Both $x^{-2}$ and $x^{4}$ are solution of the incomplete equation and they are also linearly independent, so that the general integral of the incomplete equation is ...

$y(x)= c_{1}\cdot x^{-2} + c_{2}\cdot x^{4}$ (2)

With little patience you find that $-\frac {1}{8} - \frac{x}{9}$ is solution of (1) , so that the general integral of (1) is...

$y(x)= c_{1}\cdot x^{-2} + c_{2}\cdot x^{4} -\frac {1}{8} - \frac{x}{9}$ (3)

Kind regards

$\chi$ $\sigma$

**i_zz_y_ill** · May 18th 2009, 06:52 AM

in the solution they used $y=x^m$ which is entirely different from m=lnx which is the equivalent of what your saying..im a little confused as to how you did $\frac{d}{dx}(\frac{1}{x}\frac{dz}{dt})$ aswell? ''.

**i_zz_y_ill** · May 18th 2009, 07:01 AM

i think the easist way would be if someone showed me how they did it with $y=x^m$

**chisigma** · May 18th 2009, 07:18 AM

You have to consider as first step the searching ot solution of the incomplete equation...

$x^{2}\cdot y^{''} - x \cdot y^{'} - 8 \cdot y = 0$ (1)

If you suppose that $x^{m}$ is solution of (1) , you find by simple substitution in (1) that that is true for $m=-2$ and $m=4$ . But they are linearly independent so that the general integral of (1) is...

$y(x) = c_{1}\cdot x^{-2} + c_{2}\cdot x^{4}$ (2)

The first step is terminated. The second step consists in searching a particular solution of the complete equation... not a very difficult task...

Kind regards

$\chi$ $\sigma$

**i_zz_y_ill** · May 18th 2009, 07:26 AM

I dont underrstand could you just show me how to do it using $y=x^m$

**TheEmptySet** · May 18th 2009, 07:35 AM

Originally Posted by i_zz_y_ill

The question is find the solution of:

$x^2\frac{dy}{dx}-x\frac{dy}{dx}-8y=1+x$

clearly the coefficients in what i would write down to be the auxilliary equation are not constants. Ive been told by my solution to use y= $x^m$ but i dont see how this gives(in the solutions) $m(m-1)-m-8=x^2-2m-8=(x-4)(x+2)0$
giving m=4 or m=-2. and the corrosponding complementary function. Does anyone have any ideas or could show me how the method actually cancels out the non constant coefficeints? thanks. oh i think you are supposed to use liebnitz?? not sure

If you use the anzats $y=x^m$ and take two derivatives we get

$y'=mx^{m-1}$ and

$y''=m(m-1)x^{m-2}$ Now if you plug this into the homogenious ODE you get

$x^2(m(m-1)x^{m-2})-x(m(x^{m-1}))-8x^m=0$

lets reduce to get

$m(m-1)x^m-mx^{m}-8x^m=0 \iff [m(m-1)-m-8]x^m=0$

Since $x^m \ne 0 \implies m(m-1)-m-8=0$

Solving we get $m^2-2m-8=0 \iff (m-4)(m+2)=0$ so

$m=4 \mbox{ or } m=-2$ Plug these back into your anzats and you have your complimentry solution

$y_c=c_1x^{4}+c_2x^{-2}$

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