This equation x (x2 - 1) y"+ (1- x?) y' +9(2x - 2)y has a regular singular points at x =
1 answer:
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9514 1404 393
Answer:
- y-intercept: (0, -6)
- x-intercepts: (-3, 0), (-1, 0), (1, 0)
Step-by-step explanation:
We notice the first pair of coefficients is the same as the last pair (with the sign changed). This means we can factor by grouping.
f(x) = (2x^3 +6x^2) -(2x +6)
f(x) = 2x^2(x +3) -2(x +3)
f(x) = 2(x^2 -1)(x +3) = 2(x -1)(x +1)(x +3)
The factors are made to be zero when x is 1, -1, or -3.
The x-intercepts are (1, 0), (-1, 0), (-3, 0).
The y-intercept is the constant, -6.
Answer:
The graph below shows the answer to
2x - 3y < 12
Also shown as
-3y < -2x + 12
Step-by-step explanation:
You can rearrange the inequality by subtracting 2x from both sides to isolate the y.
You now have -3y < 12 -2x
which can be put into the standard linear equation form of
-3y < -2x + 12
Then you divide both sides by -3 to get singular value of y, which is something like
-3/-3y < -2/-3x + 12/-3
which is
y > 2/3x -4
Note: I switched direction of the inequality because you are dividing both sides by a negative value.
