Question

Transform the given initial value problem into an initial value problem for two first order equations.
5. u′′ +0.25u′ +4u=2cos3t, u(0)=1, u′(0)=−2

Please haaalp

Answer 1

Typically, this sort of problem is solved by finding a total solution, which would be the sum of a homogeneous solution and a particular solution.
To find the homogeneous solution, replace the right side of the given d. e. with 0 (zero) and solve the resulting homogeneous equation. Use an auxiliary equation.
To find the particular solution, hypothesize a solution of the form A*cos 3t + B*sin 35.
Find the first and second derivactives of this. Then substitute all three (y, y' and y'') back into the original d. e., and then determine the coefficients A and B.
Hope this helps you get started.

Answer 2

I do realize I haven't discussed a transformation of this second order differential equation into two first order d. e.'s. Perhaps someone else out there is familiar with this procedure and could tell us about it. Or perhaps your textbook has an example.

Answer 3

Let x1 = u then x1' = u'

and x2 = u' then x2' = u''

As such

u(0) = 1 then x1(0) = 1

u'(0) = -2 then x2(0) = -2


So you get in total

x1' =x2

x2' = -4x1 - .25x2 +2cos(3t)

x1(0) = 1

x2(0) = -2

Hope this helps :)

Answer 4

Yes thank s.cat, way easier than I was making it, I was confused about what to do with the initial conditions but I see they can easily be expressed in terms of x1 and x2.

Answer 5

Yep :)

Answer 6

Your, just separating u into multiple equivalent variables to create a system.