The next step is to solve the recurrence, but let's back up a bit. You should have found that the ODE in terms of the power series expansion for


which indeed gives the recurrence you found,

but in order to get anywhere with this, you need at least three initial conditions. The constant term tells you that

, and substituting this into the recurrence, you find that

for all

.
Next, the linear term tells you that

, or

.
Now, if

is the first term in the sequence, then by the recurrence you have



and so on, such that

for all

.
Finally, the quadratic term gives

, or

. Then by the recurrence,




and so on, such that

for all

.
Now, the solution was proposed to be

so the general solution would be


Answer:
(3 square root of 2 , 135°), (-3 square root of 2 , 315°)
Step-by-step explanation:
Hello!
We need to determine two pairs of polar coordinates for the point (3, -3) with 0°≤ θ < 360°.
We know that the polar coordinate system is a two-dimensional coordinate. The two dimensions are:
- The radial coordinate which is often denoted by r.
- The angular coordinate by θ.
So we need to find r and θ. So we know that:
(1)
x = rcos(θ) (2)
x = rsin(θ) (3)
From the statement we know that (x, y) = (3, -3).
Using the equation (1) we find that:

Using the equations (2) and (3) we find that:
3 = rcos(θ)
-3 = rsin(θ)
Solving the system of equations:
θ= -45
Then:
r = 3\sqrt{2}[/tex]
θ= -45 or 315
Notice that there are two feasible angles, they both have a tangent of -1. The X will take the positive value, and Y the negative one.
So, the solution is:
(3 square root of 2 , 135°), (-3 square root of 2 , 315°)
Answer:
(x, y) =(-1, - 1)
Step-by-step explanation:
-Substitute the value of x
-solve the equation
-Substitute the value of y
-check the possible solution
-check solution
-simplify
Answer:it is 2
Step-by-step explanation: