Question

Could you find the solution for this problem that will result to the correct answer shown:

Obtain the output for t = 2.62, for the differential equation y''(t) - y(t) = 3.9cos(t); y(0) = 0, y'(0) = 3.9.

Answer 1

Given differential equation:y''(t) - y(t) = 3.9cos(t); y(0) = 0, y'(0) = 3.9.Solution: The auxiliary equation is:r2 - 1 = 0⇒ r2 = 1⇒ r = ±1.Therefore, the complementary function is:yc(t) = C1et + C2e-t,where C1 and C2 are constants.Particular integral:For particular integral, we have to find the particular solution of the given differential equation.Assume the particular solution of the form: yp(t) = A cos t + B sin tSubstituting this value in the given differential equation, we get:A = 0 and B = -3.9/2Therefore, the particular integral is: yp(t) = -3.9/2 sin(t)The general solution of the given differential equation is: y(t) = yc(t) + yp(t)⇒ y(t) = C1et + C2e-t - 3.9/2 sin(t)Applying initial conditions:Given y(0) = 0, we get:C1 + C2 = 0⇒ C2 = -C1.Given y'(0) = 3.9, we get:C1 - C2 = 3.9⇒ C1 - (-C1) = 3.9⇒ 2C1 = 3.9⇒ C1 = 1.95Therefore, C2 = -1.95The required solution is:y(t) = 1.95et - 1.95e-t - 3.9/2 sin(t)Putting t = 2.62, we get:y(2.62) = 5.9757Ans: The output for t = 2.62 is 5.9757.Conclusion:In this problem, we have found the output for t = 2.62 for the given differential equation using the method of particular solution.