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Problem #6: Let f(t) = { 4 0 < t < ? -4 ? < t < 2? and assume that when f(t) is extended to the negative t-axis in a periodic manner, the resulting function is even. Consider the following differential equation. 9 d^2x/dt^2 + 5x = f(t) Find a particular solution of the above differential equation of the form x_p(t) = ?_{n=1}^? A_n cos(n? t / p) = ?_{n=1}^? g(t,n) and enter the function g(t,n) into the answer box below.

          Problem #6: Let

f(t) = { 4        0 < t < ?
       -4       ? < t < 2?

and assume that when f(t) is extended to the negative t-axis in a periodic manner, the resulting function is even.

Consider the following differential equation.

9 d^2x/dt^2 + 5x = f(t)

Find a particular solution of the above differential equation of the form

x_p(t) = ?_{n=1}^? A_n cos(n? t / p) = ?_{n=1}^? g(t,n)

and enter the function g(t,n) into the answer box below.
        
Show more…
Problem #6: Let

f(t) =  4        0 < t < ?
       -4       ? < t < 2?

and assume that when f(t) is extended to the negative t-axis in a periodic manner, the resulting function is even.

Consider the following differential equation.

9 d^2x/dt^2 + 5x = f(t)

Find a particular solution of the above differential equation of the form

xp(t) = ?n=1^? An cos(n? t / p) = ?n=1^? g(t,n)

and enter the function g(t,n) into the answer box below.

Added by Daniel L.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Transcript

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00:01 In this question given function is f of t is equals to 4 minus 4 when it is 4 then the value of t is 0 to pi and when it is t then the value is 5 to 2 pi let f of t is equal to a not divided by 2 plus summation n is equal to 1 to infinity a n calls n pi t divided by p plus b n sine n pi t divided by p.
00:40 Since f of t is an equation function, so f of t is here sign term is deleted, then here a naught is equals to 2 divided by p.
00:57 Integration 0 to p f of t with respect to t where p is equal to 2 5 simplify this then we will get here the value of a knot is 0 further simplify and substitute the value then here is 10 multiplied with summation n is equal to 1 to infinity minus n squared divided by 4 a n calls n p divided by 2 plus 3 summation n is equal to 1 to infinity a n cause n pi divided by 2 simplify this then we will get here the value of a n is equal to 32 divided by n pi sine n pi divided by n pi then we will get here f of t, f of t is equals to summation n is equal to 1 to infinity 16 divided by n pi, sine n pi divided by 2, cause n t divided by 2.
02:22 Let xp of is equal to summation n is equal to summation n is equal to infinity a n .m...
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