Let

. Then

and

are two fundamental, linearly independent solution that satisfy


Note that

, so that

. Adding

doesn't change this, since

.
So if we suppose

then substituting

would give

To make sure everything cancels out, multiply the second degree term by

, so that

Then if

, we get

as desired. So one possible ODE would be

(See "Euler-Cauchy equation" for more info)
Answer: okay so trying adding each part and count how many hours is 6 am and add and I hope you get it right if not I’m sorry for you- but you should find a more sure answer
Answer:
The number is 3
Step-by-step explanation:
Hi,
4 ( 3 + 5x) = 72
12 + 20x = 72
20x = 60
x = 3
Hope this helps :)
I dont think that it belongs in primary consumers, secondary or third level consumers
Answer:
y-intercept: (0, -2)
Step-by-step explanation:
Determine by how much x (the run) and y (the rise) change as you go from (-4, -6) to (3, 1): The change in x is 7 and that in y is also 7. Thus, the slope is
m = 7/7, or m = 1.
Using the point-slope form and the point (3, 1), we get:
y - 1 = 1(x - 3), or y = 1 + x - 3, or y = x - 2
Comparing this to y = mx + b, we see that the y-intercept is -2.