If y(t) is such that dy/dt = 13(t^3)(y^2), y(0) = -4, then compute y(1).
i know your suppose to switch it up such that dy/(y^2) = (t^3)dt ... and then you take the integral of both side and get (-)1/y = 1/4(t^4) + c.... and then its this part where i screw up on the correct algebra per say, need help on the last couple of steps
You forgot that '13'
Seperating gives : $\displaystyle \displaystyle \frac{dy}{y^2} = 13 \, t^3 \, dt $
Integrating gives : $\displaystyle \displaystyle \frac{-1}{y} = \frac{13}{4} \, t^4 + C$ ... (1)
I think you can solve (1) for y. After solving, you are given that y(0)=-4 to find the value of C.
Once you have the value of C, you can calculate y(1).
