Question

Solve the initial value problem. dy/dt = 1 + 6/t , t > 0, y = 8 when t = 1

Answer 1

$\displaystyle\frac{dy}{dt} = 1 + \frac{6}{t}\ \Rightarrow\ dy = \left( 1 + \frac{6}{t}\right) dt\ \Rightarrow\int 1\, dy = \int \left( 1 + \frac{6}{t}\right) dt \ \Rightarrow \\ \\ y = t + 6\ln|t| + C. \text{ But t\ \textgreater \ 0 so }y = t + 6\ln t + C. \\ \\ y(1) = 8\ \Rightarrow\ 8 = 1 + 6 \ln 1 + C \ \Rightarrow\ C = 7 \text{ so } \\ \\ y(t) = t + 6\ln t + 7$