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 have to verify that the integral $$y(x) = \int_0^\infty \exp\left(-t - \frac{x}{\sqrt{t}}\right) dt$$ satisfies the ODE $$xy''' + 2y = 0$$ ($x > 0$). Differentiating under the integral sign three times gives $$y''' = \int_0^\infty \frac{1}{t \sqrt{t}}\exp\left(-t - \frac{x}{\sqrt{t}}\right)$$ In this form it's not obvious that this satisifes the above equation. I guess I need to use integration by parts to simplify the above integral but I can't quite see how that works. If I let $u = \exp(-t - \frac{x}{\sqrt{t}})$ and $v' = \frac{1}{t \sqrt{t}}$ then $v = \frac{-2}{\sqrt{t}}$ and $u' = \left(-1 + \frac{x}{2t\sqrt{t}}\right)\exp(-t - \frac{x}{\sqrt{t}})$ and the above becomes $$y''' = \int_0^\infty \frac{-2}{\sqrt{t}} \left(-1 + \frac{x}{2t\sqrt{t}}\right)\exp\left(-t - \frac{x}{\sqrt{t}}\right) \, dt$$ and this doesn't seem to work.

Can anyone provide me with help? Hopefully I'm not just making a mistake with the differentiation/integration.

share|improve this question
    
I changed \textrm{exp} to \exp. That not only prevents italicization but also provides proper spacing in expressions like $a\exp b$. That is standard usage. –  Michael Hardy Aug 7 '13 at 23:40
1  
Why would you differentiate three times? –  Pedro Tamaroff Aug 7 '13 at 23:44
1  
Should the equation be $xy'''+2y=0$? –  Pedro Tamaroff Aug 7 '13 at 23:49
    
Yes, typo, sorry –  rt93 Aug 7 '13 at 23:51
add comment

1 Answer

up vote 5 down vote accepted

We have that $$y'(x) = - \int_0^\infty {\frac{1}{{\sqrt t }}\exp } \left( { - t - \frac{x}{{\sqrt t }}} \right)dt$$

and then $$y'''(x) = -\int_0^\infty {\frac{1}{t\sqrt t}\exp } \left( { - t - \frac{x}{{\sqrt t }}} \right)dt$$

Now, note that then we have $$xy''' + 2y = \int_0^\infty {\left( {2 - \frac{x}{{t\sqrt t }}} \right)\exp \left( { - t - \frac{x}{{\sqrt t }}} \right)dt} $$

Now, write this as $$xy''' + 2y = - 2\int_0^\infty {\left( {-1 + \frac{x}{{2t\sqrt t }}} \right)\exp \left( { - t - \frac{x}{{\sqrt t }}} \right)dt} $$ and note that$$\frac{d}{{dt}}\left( { - t - \frac{x}{{\sqrt t }}} \right) = - 1 + \frac{x}{2{t\sqrt t }}$$

Can you continue? Note that the integration bounds should collapse since $$\eqalign{ & \mathop {\lim }\limits_{t \to {0^ + }} \left( { - t - \frac{x}{{\sqrt t }}} \right) = - \infty \cr & \mathop {\lim }\limits_{t \to \infty } \left( { - t - \frac{x}{{\sqrt t }}} \right) = - \infty \cr} $$

ADD If you want to make this a little more clear, split at the point where the function is zero, namely $$\int_0^{{{\left( {\frac{x}{2}} \right)}^{2/3}}} {\left( { - 1 + \frac{x}{{2t\sqrt t }}} \right)\exp \left( { - t - \frac{x}{{\sqrt t }}} \right)dt} + \int_{{{\left( {\frac{x}{2}} \right)}^{2/3}}}^\infty {\left( { - 1 + \frac{x}{{2t\sqrt t }}} \right)\exp \left( { - t - \frac{x}{{\sqrt t }}} \right)dt} $$

Then we have under the same substitution this is $$\int_{ - \infty }^0 {\exp udu} + \int_0^{ - \infty } {\exp udu} = 0$$

share|improve this answer
    
I see now. Thanks. So integration by parts wasn't necessary after all... –  rt93 Aug 7 '13 at 23:58
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.