hw1sol

# hw1sol - Solutions for Homework 1 1 2.1.1 Fixed points are...

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Unformatted text preview: Solutions for Homework 1 1. 2.1.1 Fixed points are points where sin x = 0, that is, integer multiples of π . 2.1.2 The greatest velocity to the right occurs where sin x is the greatest. That is, where x is π / 2 plus an integer multiple of 2 π . 2.1.3 a) ˙ x = sin x implies ¨ x = ( cos x ) ˙ x by the chain rule, so ¨ x = cos x sin x = sin ( 2 x ) / 2. Did you remember that cos x sin x = sin ( 2 x ) / 2? If not, here is an easy way to memorize it: Among the trigonometric formulas, the only ones that I memorize are sin ( a + b ) = sin a cos b + cos a sin b and cos ( a + b )= cos a cos b − sin a sin b . All others follow from these very quickly. For example, the first with a = b = x gives you what you need for this problem. b) The acceleration to the right is greatest where sin ( 2 x ) is greatest, that is where x = π / 4 plus an integer multiple of π . 2. 2.2.1. Sketch of the vector field:-4-2 2 4-20-10 10 20 30 40 x 4 x 2-16 x = − 2 is a stable fixed point, and x = 2 is unstable, as is apparent from the vector field. Here is what the solutions look like: 0.1 0.2 0.3 0.4-3-2-1 1 2 3 t x (Solutions with x > 2 blow up in finite time.) Solution in closed form: If x = ± 2, then x ( t ) = ± 2 for all t , since 2 and − 2 are fixed points. Let’s assume that x negationslash = ± 2 now. Then x ( t ) negationslash = ± 2 for all t (this is true by the uniqueness part of 1 the existence and uniqueness theorem). This will be used in the following manipulations. ˙ x = 4 x 2 − 16 ⇔ ˙ x x 2 − 4 = 4 ⇔ 4 ˙ x ( x − 2 )( x + 2 ) = 16 ⇔ ˙ x x − 2 − ˙ x x + 2 = 16 ⇔ d dt ln ( | x − 2 | ) − d dt ln ( | x + 2 | ) = 16 ⇔ d dt ln parenleftbiggvextendsingle vextendsingle vextendsingle vextendsingle x − 2 x + 2 vextendsingle vextendsingle vextendsingle vextendsingle parenrightbigg = 16 ⇔ ln parenleftbiggvextendsingle vextendsingle vextendsingle vextendsingle x − 2 x + 2 vextendsingle vextendsingle vextendsingle vextendsingle parenrightbigg = 16 t + C for some constant C , which is found by plugging in t = 0. Using the notation x ( ) = x , we find C = ln parenleftbiggvextendsingle vextendsingle vextendsingle vextendsingle x − 2 x + 2 vextendsingle vextendsingle...
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hw1sol - Solutions for Homework 1 1 2.1.1 Fixed points are...

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