Question

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).

Answer 1

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.