What is the trial particular solution for this differential equation?

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SUMMARY

The discussion focuses on finding the trial particular solution for the differential equation y'' + 24y' + 432y = cos(wt). The proposed trial solution is yp(x) = R(h) cos(ht - ∅(h)), with the objective to determine R(h) and the value hbar, where R(h) is maximized. The value of hbar is established as 12, and the approach involves solving the coefficients for sin(Ht) and cos(Ht) using the quadratic equation derived from the differential equation.

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  • Knowledge of trigonometric identities, particularly the double angle formula for cosine.
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dan5
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Homework Statement



y'' + 24y' +432y = cos (wt)

yp(x) = R(h) cos (ht - ∅(h))

find R(h)
and hbar = value for h which R(h) is max.
and ∅(hbar)

Homework Equations



double angle formula for cos
cos (a-b) = cos a cos b + sin a sin b

The Attempt at a Solution



solving the equation as a quadratic in coefficients on sin and cos, hbar = 12
 
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Frankly, I can't figure out what your question is. What is yp? Is it supposed to be a particular solution? If so, you will need to find yp' and yp'' and put them into the equation. What do you get when you do that?
 
yes yp is the particular solution.

Well i thought of using a trial particular solution of

yp= A sin (Ht) + B cos (Ht)

which leaves me with

-AH2-24BH+432A = 0 for coefficients of sin(Ht)

and

-BH2+24AH+432B = 1 for coefficients of cos(Ht)
 

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