That is a "linear equation" so you can find the general solution to the associated homogeneous equation, x'= -\alpha_kt x, and a single solution to the entire equation and add them.
Of course, x'= -\alpha_k tx "separates" into
\frac{dx}{x}= -\alpha_k t dt[/itex]<br />
which can be easily integrated to give <br />
x= Ae^{-\alpha_k t^2/2<br />
<br />
To find a single solution to the entire equation, look for a solution of the form<br />
x(t)= u(t)e^{-\alpha_k t^2/2}[/itex]<br />
so that <br />
x&amp;#039;= u&amp;#039; e^{-\alpha_k t^2/2}- \alpha_k t e{-\alpha_k t^2/2}[/itex]&lt;br /&gt;
and putting that into the differential equation gives&lt;br /&gt;
u&amp;amp;#039; e^{-\alpha_k t^2/2}- \alpha_k t u e{-\alpha_k t^2/2}= -\alpha_k t u e^{-\alpha_k t^2/2}+ Ce^{i(\alpha_k- \alpha_m)t^2&lt;br /&gt;
&lt;br /&gt;
The &lt;br /&gt;
-\alpha_k t u e^{-\alpha_k t^2}[/itex] &amp;lt;br /&amp;gt;
terms cancel leaving &amp;lt;br /&amp;gt;
u&amp;amp;amp;#039;e^{-\alpha_k t^2/2}= Ce^{i(\alpha_k- \alpha_m)t^2&amp;lt;br /&amp;gt;
so you can now solve for u(t).
