Question

Solve the initial value problem explicitly: dx/dt = (8/t) - (1/t^4) + 8 and x = 6 when t = 1

Answer 1

what are you solving for and is this the entire question?

Answer 2

Yes i have to find the integral of that equation and find C

Answer 3

Any help?

Answer 4

\[dx=\left[ \frac{ 8 }{ t }-\frac{ 1 }{ t^4 }+8 \right]dt\]
integrate both sides
\[x=8 \int\limits \frac{ 1 }{ t }dt- \int\limits t ^{-4}dt+8\int\limits dt+c\]
\[x=8 \ln \left| t \right|-\frac{ t ^{-3} }{ -3 }+8t+c\]
find c when x=6,t=1
then substitute the value of c

Answer 5

How do you know t^-4 integrates to t^-3/-3?

Answer 6

I got C = -7/3

Answer 7

\[\int\limits x^n dx =\frac{ x ^{n+1} }{ n+1 }\]

Answer 8

\[\int\limits t^{-4}dt=\frac{ t ^{-4+1} }{ -4+1 }=\frac{ t ^{-3} }{ -3 }\]

Answer 9

So would the integral of -(24/x^7) = x^-6/-144?

Answer 10

\[\int\limits \frac{- 24 }{ x^7 }dx=-24*\frac{ x ^{-6} }{ -6 }\]
\[=\frac{ 4 }{ x^6 }\]

Answer 11

Why does the variable move to the bottom?