Extraordinary Differential Equations #12

Calculus Level 4

Consider the differential equation on a series of cosine functions involving y = y ( x ) y=y(x) : d y d x = k = 1 2017 k cos ( ( 2 k 1 ) y ) \frac{dy}{dx}=\sum_{k=1}^{2017} k\cos{((2k-1)y)} where y ( 0 ) = π 2 y(0)=\dfrac{\pi}{2} . Determine the value of y ( 2017 ) π \dfrac{y(2017)}{\pi} .


This problem is part of the set Extraordinary Differential Equations .
1 3 \frac{1}{3} 0 1 2017 \frac{1}{2017} 1 π \frac{1}{\pi} 1 2 \frac{1}{\sqrt{2}} 1 1 4 \frac{1}{4} 1 2 \frac{1}{2}

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1 solution

Wee Xian Bin
Feb 26, 2017

Notice that if there is an equilibrium solution to the ODE, y ( x ) = π 2 y(x)=\frac{\pi}{2} for all x x , whereby y ( x ) = d d x π 2 = 0 y'(x)=\frac{d}{dx} \frac{\pi}{2}=0 for all x x , y ( x ) = k = 1 2017 k cos ( ( 2 k 1 ) ( π 2 ) ) = 0 y'(x)=\sum_{k=1}^{2017} k\cos{((2k-1)(\frac{\pi}{2}))=0} for all x x , and this also satisfies the initial condition y ( 0 ) = π 2 y(0)=\frac{\pi}{2} . Thus, y ( 2017 ) π = π 2 × 1 π = 1 2 \frac{y(2017)}{\pi}=\frac{\pi}{2}\times\frac{1}{\pi}=\frac{1}{2} .

How do you know there is no other solution ?

Kushal Bose - 4 years, 3 months ago

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