Solving for x(t) in a Nonlinear Differential Equation - Expert Help Needed

[Amok]

Hey guys, can you help me to find a solution to this?

x'(t) = -\alpha_k t x(t) + Ce^{i(\alpha_k - \alpha_m)t^2}

Where C and the alphas are known constants and the initial condition is that x(0) equals some constant. I have no idea on where to start.

 

[HallsofIvy]

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]&lt;br /&gt; so that &lt;br /&gt; x&amp;amp;#039;= u&amp;amp;#039; e^{-\alpha_k t^2/2}- \alpha_k t e{-\alpha_k t^2/2}[/itex]&amp;lt;br /&amp;gt; and putting that into the differential equation gives&amp;lt;br /&amp;gt; u&amp;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&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; The &amp;lt;br /&amp;gt; -\alpha_k t u e^{-\alpha_k t^2}[/itex] &amp;amp;lt;br /&amp;amp;gt; terms cancel leaving &amp;amp;lt;br /&amp;amp;gt; u&amp;amp;amp;amp;#039;e^{-\alpha_k t^2/2}= Ce^{i(\alpha_k- \alpha_m)t^2&amp;amp;lt;br /&amp;amp;gt; so you can now solve for u(t).

 

[Amok]

Actually, I knew how to do that. It was pure intellectual laziness (variation of constants), shame on me. Thanks HOI.