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20 February, 04:01

If y1 (t) is a particular solution to 3y'' - 5y' + 9y = te^t and y2 (t) is a particular solution to 3y'' - 5y' + 9y = tan (3t), then what is the differential equation that has a particular solution of y2 (t) - 5y1 (t) ? * Hint: Super Position Principle

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  1. 20 February, 04:20
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    Step-by-step explanation:

    We will use the superposition principle. REcall that this principle says that if the particular function f on the side that doesn't depend on y could be written as a sum of functions (say f=g+h). Then, you can solve the equation by solving for g and solving for h and then summing up the solutions.

    Consider a linear differential equation of the form ay''+by'+cy+d = f (t) (a, b, c, d are constants) with particular solution y1 (t) and the differential equation ay''+by'+cy+d = g (t) with particular solution y2 (t). So if we have the equation

    ay''+by'+cy+d = kf (t) + hg (t), then the particular solution will be ky1 (t) + hy2 (t).

    Hence since we want y2 (t) - 5y1 (t) to be the solution, we identify k=-5 and h=1. So the differential equation whose particular solution is the desired is the following

    3y''-5y'+9y = - 5te^t+tan (3t)
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