Answer:
7 yards
Step-by-step explanation:
gained 6 yards on a first down +6
lost 15 yards on the second down - 15
gained 12 yards on the third down +12
At the end of the third down
+6 -15+12 = +3
They want to have a 10 yard gain at the end of the 4th down
+3 + x = +10
Subtract 3 from each side
+3-3 +x = +10-3
x = 7
They need to gain 7 yards
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

Lmk if you need more help :)
Answer:
The first graph would be correct
Step-by-step explanation: