second order differential equations again

**djmccabie** · May 17th 2009, 06:00 AM

A particle moves along the x-axis so that the time ts, its displacement xm from the origin satisfies the differential equation

$\frac{d^2x}{dt^2}+2\frac{dx}{dt}-35x=70t + 31$

Given that when t=0, the particle is at rest at the origin, find its displacement at time ts.

I'm struggling with this because i think it involves complex roots, which i haven't done before.i know the C. Function is $e^{\alpha t}(Acos\beta t + \beta sin \beta t)$

and from the equation i get

$d^2 + 2d - 35$

$\frac{-2^+_- \sqrt{4-140}}{2}$

$\frac{-2^+_- \sqrt{-136}}{2}$

Im a bit stuck and where to go from here. Can i still use surds when its complex? ie $\sqrt{-136}=2\sqrt{-34}$ then i can cancel the 2's with the denominator?

**Scott H** · May 17th 2009, 06:25 AM

You are correct that

$d^2+2d-35=0$

gives us the auxiliary roots for the homogenous solution of $x''+2x'-35x=70t+31$ . To solve this, we factor it into

$(d+7)(d-5)=0.$

Using the quadratic formula, we obtain the same answers:

$\frac{-2\pm\sqrt{2^2-4\cdot1\cdot(-35)}}{2}=\frac{-2\pm\sqrt{144}}{2}=\frac{-2\pm 12}{2}=-1\pm 6=-7\,\mbox{or}\,5.$

It can be shown that the general solution is of the form

$x=x_h+x_p$

where $x_h$ is the homogeneous solution and $x_p$ is a particular solution of $x''+2x'-35x=70t+31$ . Since $70t+31$ is a first-degree polynomial, we may try

$x_p=A t+B$

as a solution and then try to figure out what the values of $A$ and $B$ might be.

**djmccabie** · May 17th 2009, 06:29 AM

OK it was just me making a silly mistake! haha thanks

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