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Radda [10]
3 years ago
8

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
Mathematics
1 answer:
jonny [76]3 years ago
6 0

Answer:

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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The graph is attached Below and the plotting is given below.

Step-by-step explanation:

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To draw a line on a graph the required minimum two points but here we will have three points as point A, point B, and point C.

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