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
[tex]\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 /> [tex]x(t)= u(t)e^{-\alpha_k t^2/2}[/itex]<br /> so that <br /> [tex]x'= u' e^{-\alpha_k t^2/2}- \alpha_k t e{-\alpha_k t^2/2}[/itex]<br /> and putting that into the differential equation gives<br /> $$u' e^{-\alpha_k t^2/2}- \alpha_k t u e{-\alpha_k t^2/2}= -\alpha_k t u e^{-\alpha_k t^2/2}+ Ce^{i(\alpha_k- \alpha_m)t^2$$<br /> <br /> The <br /> [tex]-\alpha_k t u e^{-\alpha_k t^2}[/itex] <br /> terms cancel leaving <br /> $$u'e^{-\alpha_k t^2/2}= Ce^{i(\alpha_k- \alpha_m)t^2$$<br /> so you can now solve for u(t).[/tex][/tex][/tex][/tex]

 

[Amok]

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