The Seventh of Seven Problems
The average of all my grades is x + 0.35(don't judge).X is where where is the solution to . Find x.B and C are constants of integration.
The answer is 98.
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1 solution
.The hard part of the problem is solving for the solution of the differential equation above.We can factor the equation as y(D+I)^3 = 0 .Rewrite the equation as z'+z=0 where z is y(D+I)^2.By basically common sense the solution is e^-t.Now set e^-t=y(D+I)^2.Rewrite as e^-t = n'+n.Multiply by e^t and you have a product loop y'e^t+ye^t.Integrate and you have ye^t = t+B and multiply out by e^-t.So te^-t +Be^-t = y'+y.Repeating the technique shown earlier we have e^-t(t^2/2+Bt+C)(note that the constants of integration are reversed in the problem).Plugging in e^-t as g(t) we have e -e^f(e).f(e)=1 so e-e=0.x= 2^6+2^5+2^1 = 98 the answer.98.35? terrible!