# Thread: I'm really stuck with this 3rd order ODE

1. ## I'm really stuck with this 3rd order ODE

I've been working on this one for about 3 hours and can't seem to find a solution for it:

$x^3y''' - 3x^2y'' +7xy' - 8y = x + exp(2x) for x > 0$

I understand it's a 3rd order Cauchy-Euler and I've solved for the characteristic solution as:

$y=c_1x^2 + c_2x^2ln(x)+c_3x^2ln(x)^2$

Now, I have no idea how to find the solution when g(x) = x + exp(2x). I've tried using the Variation of Parameters technique only to arrive at ugly integrals. I'm assuming you have to make some sort of substitution.

Any help, much appreciated. Thanks.

2. there's actually a faster way to do this, it's applicable on superior ODEs that involves matrices and other stuff.

haven't you ever covered?

3. Originally Posted by AKTilted
I've been working on this one for about 3 hours and can't seem to find a solution for it:

$x^3y''' - 3x^2y'' +7xy' - 8y = x + exp(2x) for x > 0$

I understand it's a 3rd order Cauchy-Euler and I've solved for the characteristic solution as:

$y=c_1x^2 + c_2x^2ln(x)+c_3x^2ln(x)^2$

Now, I have no idea how to find the solution when g(x) = x + exp(2x). I've tried using the Variation of Parameters technique only to arrive at ugly integrals. I'm assuming you have to make some sort of substitution.

Any help, much appreciated. Thanks.
With the substitution $\ln x = t$ Your DE becomes...

$\displaystyle y^{'''} (t) -4\ y^{''} (t) +12\ y^{'} (t) -8\ y(t) = e^{t} + e^{(2 e^{t})}$ (1)

... that is 'linear homogeneous with constant coefficients' so that the search of the 'particular solution' due to the 'known term' $t^{2} + \ln t$ can be performed in 'standard procedure'...

Kind regards

$\chi$ $\sigma$