Solve the following initial value problem: with
So I isolated the y and it became
8y=(9-dy/dt)t
how would I finish the rest?
You can also do this as a "homogeneous" equation. Write it as $\displaystyle \frac{dy}{dt}= \frac{9t- 8y}{t}= 9- 8\frac{y}{t}$.
Let $\displaystyle u= \frac{y}{t}$. Then $\displaystyle y= tu$ so y'= tu'+ u and the equation becomes tu'+ u= 9- 8u and so tu'= 9- 9u, a separable equation.
$\displaystyle \frac{dy}{dt} + \frac{8}{t}y = 9$.
The Integrating Factor is $\displaystyle e^{\int{\frac{8}{t}\,dt}} = e^{8\ln{t}}$
$\displaystyle = e^{\ln{t^8}}$
$\displaystyle = t^8$.
So multiply both sides by the integrating factor to get
$\displaystyle t^8\left(\frac{dy}{dt} + \frac{8}{t}y\right) = 9t^8$
$\displaystyle t^8\,\frac{dy}{dt} + 8t^7y = 9t^8$
Can you see that the LHS is a product rule expansion of $\displaystyle \frac{d}{dt}(t^8y)$?
$\displaystyle \frac{d}{dt}(t^8y) = 9t^8$
$\displaystyle t^8y = \int{9t^8\,dt}$
$\displaystyle t^8y = t^9 + C$
$\displaystyle y = t + Ct^{-8}$.
Now use the initial condition to find $\displaystyle C$.
It helps if you write the DE as
$\displaystyle \frac{1}{1 - u}\,\frac{du}{dt} = \frac{9}{t}$
so that when you take the integral of both sides with respect to t
$\displaystyle \int{\frac{1}{1 - u}\,\frac{du}{dt}\,dt} = \int{\frac{9}{t}\,dt}$
$\displaystyle \int{\frac{1}{1 - u}\,du} = \int{\frac{9}{t}\,dt}$.
I'm sure you can go from here.