Help Entering Answers See Example 4.1.4, in Section 4.1, in the MTH 235 Lecture

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Help Entering Answers See Example 4.1.4, in Section 4.1, in the MTH 235 Lecture Notes (10 points) Consider the initial value problem (a) Write the differential equation above as a first order system for the unknowns t, t, where u = y You do not need to solve the system. Note: Type u for u(t) and v for v(t) (b) Write the initial conditions for the functions u, v a(0) = v(0) Note: You can earn partial credit on this probiem Show me another Preview My Answers Submit Answers You have attempted this problem 0 times You have 20 attempts remaining e to search

Explanation / Answer

The question already tells us u = y.
What we do for this kind of problem is to set the other unknown, v, as v = y'

So we have:
u = y
v = y'
Consequently it ought to be clear that:
u' = y'
v' = y''

Notice from the above that because v = y' and u' = y' then u' = v

Initial conditions:
If u = y, because y(0) = -1, then u(0) = -1
If v = y', because y'(0) = -4, then v(0) = -4

Now then, from the second order equation, rearrange for y''. We get:

y''= 6y'-5y+4sin(4t)

Putting all of the above together, we get:

u' = v

v' = 6v-5u+4sin(4t)

with

v(0) = -4

u(0) = -1

I hope this answer helps.
Thanks.

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