How do I solve this Differential Equation in closed form.

Question

sh3lsh · Answer

$$x^2y ^{''}(x)+xy ^{'}(x)-y(x)=e ^{x-1}, y(1)=1, y ^{'}(1) = 0$$

dogzcatz · Answer

Do you know how to solve a Differential Equation

sh3lsh · Answer

Yeah! I just wanted some help getting started. Should I use Laplace for this?

sh3lsh · Answer

To answer my own question, no. 
However, this g(t) messes me up.

sh3lsh · Answer

@ganeshie8 Would you mind assisting for a bit?

zepdrix · Answer

Hmm I'm a little rusty on my diff eq...
is this a Cauchy-Euler equation though? :)

zepdrix · Answer

Solutions will be of the form: $\large\rm y=x^m$
instead of the exponentials that we're used to.

Sound familiar or no? :)

sh3lsh · Answer

Hmm. I don't suppose I've learned strictly about this equation! 
Let me read about this for a bit, and I'll get back to you. 
In the meanwhile, I'll close this question! 

Thanks!

irishboy123 · Answer

it's E-C when homogeneous

so the complementary solution is a shoo-in.   use zep's sub or $x = e^t$ is really good

so follow up with var of params for the particular solution?

if you go to length of doing it, please share!  it crashes Wolfram...for me at least

i reckon Laplace is a non starter.  try finding the transforms you need, even if you re-org the equation a bit.

irishboy123 · Answer

@freckles @SithsAndGiggles

anonymous · Answer

Replacing $x$ with $e^t$ is a good start. Let $z(t)$ be the target function, where $t=\ln x$ or $x=e^t$. Then $x\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{\mathrm{d}z}{\mathrm{d}t}$ and $x^2\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=\dfrac{\mathrm{d}^2z}{\mathrm{d}t^2}-\dfrac{\mathrm{d}z}{\mathrm{d}t}$. Under this substitution the ODE becomes $$z''-z=e^{e^t-1}$$with initial conditions $z(0)=1$ and $z'(0)=0$. The Laplace transform doesn't work in this situation because the RHS is not of exponential order (meaning there's no constant $M$ such that $\left|e^{e^t-1}\right|\le Me^t$). Variation of parameters or reduction of order both seem to work, but arriving at a solution would involve some familiarity with writing the solution(s) in terms of definite integrals, or this special function: http://mathworld.wolfram.com/ExponentialIntegral.html