# Need some quick help with differential equations...

**URL:** <https://boards.straightdope.com/t/need-some-quick-help-with-differential-equations/299408>\
**Category:** Factual Questions\
**Created:** [April 15, 2005, 1:37am UTC](https://boards.straightdope.com/t/need-some-quick-help-with-differential-equations/299408 "2005-04-15T01:37:07Z")\
**Posts on this page:** 1\
**Page:** 1

<div class="post-metadata">

**Author:** ![BabaBooey](https://avatars.discourse-cdn.com/v4/letter/b/dfb087/32.png) [@BabaBooey](https://boards.straightdope.com/u/BabaBooey)\
**Post date:** [April 15, 2005, 1:37am UTC](https://boards.straightdope.com/t/need-some-quick-help-with-differential-equations/299408/1 "2005-04-15T01:37:07Z")

</div>

Teacher has an old test up for review for an exam tomorrow. One problem:

•2. (a) Solve DE x’ ’ + 4x = 0 with the initial conditions x(0) = 2 and x’(0) = -2

This is simple harmonic motion and the general solution is x(t) = c1 cos(2t) + c2 sin(2t), with  
x’(t) = -2_c1 sin(2t) + 2_c2 cos(2t) and the initial conditions yield

x(0) = 2 = c1  
x’(0) = -2 = 2 c2 or c2 = -1

So the solution to the IVP is x(t) = 2 cos(2t) - sin(2t)

I understand all of that, part b though…

• b) Express the solution in phase amplitude form (it’s an image, so my writing it will look a bit awkward) :

A = [(2)^3 + (-1)^3]^ (1/2)

Where exactly did the ^3 come from? I was under the impression that A = the square root of x(t)^2 + (x’(t)/w)^2. I apoligize for the incredibly boring question.