# Integral solution of a differential equation (verification)

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.

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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
Why would you differentiate three times? –  Pedro Tamaroff Aug 7 '13 at 23:44
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

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 $$\int_{ - \infty }^0 {\exp udu} + \int_0^{ - \infty } {\exp udu} = 0$$

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