You must be dumb and have no friends
Answer:I feel bad for you
I don't know what method is referred to in "section 4.3", but I'll suppose it's reduction of order and use that to find the exact solution. Take

, so that

and we're left with the ODE linear in

:

Now suppose

has a power series expansion



Then the ODE can be written as


![\displaystyle\sum_{n\ge2}\bigg[n(n-1)a_n-(n-1)a_{n-1}\bigg]x^{n-2}=0](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bn%5Cge2%7D%5Cbigg%5Bn%28n-1%29a_n-%28n-1%29a_%7Bn-1%7D%5Cbigg%5Dx%5E%7Bn-2%7D%3D0)
All the coefficients of the series vanish, and setting

in the power series forms for

and

tell us that

and

, so we get the recurrence

We can solve explicitly for

quite easily:

and so on. Continuing in this way we end up with

so that the solution to the ODE is

We also require the solution to satisfy

, which we can do easily by adding and subtracting a constant as needed:
Answer:
i really dont know how to do this and i dont think anybody here knows how to
Step-by-step explanation:
Substitute y=4x to the second equation:
x^2 + (4x)^2 = 17
x^2 + 16x^2 = 17
17x^2 = 17
x^2 = 17/17
x^2 = 1
x = 1 and -1
When x=1, y=4(1) = 4
When x=-1, y=4(-1) = -4
Thus the solutions would be (1,4) and (-1,-4). That would correspond to D. and A.
Answer:
75%
Step-by-step explanation:
Happy Friday!