A homogeneous second-order linear differential equation, two functions y1 and y2 , and a pair of initial conditions are given below. First verify that y1 and y2 are solutions of the differential equation. Then find a particular solution of the form y = c1y1 + c2y2 that satisfies the given initial conditions.
y'' + 49y = 0; y1 = cos(7x) y2 = sin(7x); y(0) = 10 y(0)=-4
y(x)=?

Step-by-step explanation:

Check part

Now, replace to the original one.

Done!!

Particular solution

I believe that y'(0) = 4, not y(0) anymore. Since y(0) CANNOT have two different solution.

The last step is to put C1, C2 into your solution. You finish it.

question 1 is 8, question 2 is 9, question 3 is 24

step-by-step explanation:

now, the numbers they give you at the top are the key. from there, you just have to plug and chug! let's try 1, 2, and 3!

key: a=3, b=5, c=6

1. a+5

remember, a=3, so we plug that in to get 3+5=8

2. 15-c

c=6, so we plug that in to get 15-6=9

3. 4b

b=6, so we plug it in to get 4*6=24

time to try the rest on your own! if you put the answers below in a comment, i'll check them for you!