Question

Solve the differential equation. dp/dt = t^2p - p + t^2 - 1

Answer 1

$\frac{dp}{dt} = t^2 p - p + t^2 - 1 \Rightarrow \frac{dp}{dt} = p(t^2 - 1)+ 1(t^2 - 1 ) \Rightarrow \\ \\ \frac{dp}{dt} = (p+1)(t^2 - 1)\ \Rightarrow\ \frac{dp}{p+1} = \left(t^2 - 1\right)dt\ \Rightarrow \\ \\ \int \frac{dp}{p+1} =\int \left(t^2 - 1\right)dt\ \Rightarrow \ \ln|p+1| = \frac{1}{3}t^2 - t + C \ \Rightarrow \\ \\ |p+1| = e^{t^3/3 - t + C}\ \Rightarrow\ p+1 = \pm e^Ce^{t^3/3 - t} \Rightarrow \\ \\ p = Ke^{t^3/3 -t} - 1,\ \text{where K = \pm e^C }$

Since $p = -1$ is also a solution, $K$ can equal 0, and hence, $K$ can be any real number.