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

So first, you get the absolute value to one side:

Next, set up two equations; One where the value inside the absolute value lines is positive, and another where it is negative, and solve both for the variable:

Your answers are
8 and -8, or +-8.
Diameter= 2 times the radius
2r=d
2r=10.2
R= 10.2/2
R=5.1
Therefore, the radius is 5.1 units which makes the answer choice C correct.
Hopefully, this helps!
Answer:
x=-2
Step-by-step explanation:
It is a straight line when x=-2 :)
13/17 if you are trying to simplify it. You divide it by 3 on the top and bottom