Recall that variation of parameters is used to solve second-order ODEs of the form
<em>y''(t)</em> + <em>p(t)</em> <em>y'(t)</em> + <em>q(t)</em> <em>y(t)</em> = <em>f(t)</em>
so the first thing you need to do is divide both sides of your equation by <em>t</em> :
<em>y''</em> + (2<em>t</em> - 1)/<em>t</em> <em>y'</em> - 2/<em>t</em> <em>y</em> = 7<em>t</em>
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You're looking for a solution of the form

where


and <em>W</em> denotes the Wronskian determinant.
Compute the Wronskian:

Then


The general solution to the ODE is

which simplifies somewhat to

2l + 2w = 132
l = 3w + 2
2(3w+2)+2w=132
6w + 4 + 2w = 132
8w + 4 = 132
8w = 128
w = 16
l = 3(16) + 2
l = 48 + 2
l = 50
2(50) + 2(16) __ 132
100 + 32 = 132
Answer:
1350π
should be your answer
Step-by-step explanation:
15 x 15 x π = 225π
225π x 6 = 1350π
Answers
1. B) 31
2. B) 4
3. A) 15
4. D) 24
5. A) 3
MARK BRAINLEIST PLS THIS TOOK A LOT OF TIME
Answer:
8
Step-by-step explanation:
We can use triangle inequality, let AC = x:
x+4 > 5
x+5 > 4
9 > x
Simplifying:
x>1
x>-1
9>x
Thus: 9>x>1
So the Largest whole number is 8