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Sagot :
Answer:
y(v) = [tex]8\ln |v| + \frac{v^{3} }{2} + \frac{ v^{2}}{2} - 12v[/tex] - 10
Step-by-step explanation:
y''(v) =[tex]3v - \frac{8}{v^{2} } + 1[/tex]
y'(v) = [tex]\frac{3v^{2} }{2} + v + \frac{8}{v}[/tex] + C
y'(2) = [tex]\frac{3(2)^{2} }{2} } + 2 + \frac{8}{2}[/tex] + C = 6 + 2 + 4 + C = 12 + C = 0
C = -12
So, y'(v) = [tex]\frac{3v^{2} }{2} + v + \frac{8}{v}[/tex] -12
Now, y(v) = [tex]8\ln |v| + \frac{v^{3} }{2} + \frac{ v^{2}}{2} - 12v[/tex] + D
Now, y(-1) = [tex]8\ln |-1| + \frac{(-1)^{3} }{2} + \frac{(-1)^{2} }{2} - 12(-1)[/tex] + D = 2
[tex]0 + \frac{-1}{2} + \frac{1}{2} +12[/tex] + D = 2
12 + D = 2
D = -10
Wow! Now y(v) = [tex]8\ln |v| + \frac{v^{3} }{2} + \frac{ v^{2}}{2} - 12v[/tex] - 10
I spent a long time typing this, so I hope it is correct for you.
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