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

Graphing is one way to do the problem.But sometimes, graphing it is hard to do.So here’s an algebraic method.
If M(m1, m2) is the midpoint of two points A(x1, y1) and B(x2, y2),then m1 = (x1 + x2)/2 and m2 = (y1 + y2)/2.In other words, the x-coordinate of the midpointis the average of the x-coordinates of the two points,and the y-coordinate of the midpointis the average of the y-coordinates of the two points.
Let B have coordinates (x2, y2) in our problem.Then we have that 6 = (2 + x2)/2 and 8 = (3 + y2)/2.
Solving for the coordinates gives x2 = 10, y2 = 13
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