142k views
3 votes
Given that y1 = 1/4 sin(2t) is a solution of y" + 2y' + 4y = cos(2t) and y2 = t/4 - 1/8 is a solution of y" + 2y' + 4y = t, use the superposition principle to find a solution of y" + 2y + 4y = 2t - 3cos(2t).

1 Answer

1 vote

Final answer:

Using the superposition principle by combining the given solutions y1 = 1/4 sin(2t) and y2 = t/4 - 1/8, the solution for the differential equation y" + 2y' + 4y = 2t - 3cos(2t) is y = t/2 - 1/4 - 3/4 sin(2t).

Step-by-step explanation:

To find a solution of y" + 2y' + 4y = 2t - 3cos(2t) using the superposition principle, we can use the given solutions y1 and y2 for the related differential equations. The superposition principle states that the sum of the solutions to a linear homogeneous differential equation and a particular solution to the non-homogeneous equation represents the general solution to the non-homogeneous equation.

In this case, we have two solutions:

  • y1 = 1/4 sin(2t), a solution of y" + 2y' + 4y = cos(2t)
  • y2 = t/4 - 1/8, a solution of y" + 2y' + 4y = t

The given equation y" + 2y' + 4y = 2t - 3cos(2t) can be seen as a linear combination of the two right-hand sides of the equations for which we already know solutions. Hence, a solution can be constructed as:

y = 2y2 - 3y1

Plugging in the known forms of y1 and y2:

y = 2(t/4 - 1/8) - 3(1/4 sin(2t))

This is the solution of the superposed differential equation, and simplifies to:

y = t/2 - 1/4 - 3/4 sin(2t)

Note that it is essential that the original differential equations were linear for this principle to apply.

User Ramu Pasupuleti
by
8.5k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories