Question

\({\bf{1 - 8}}\)Solve the differential equation.

\(\frac{{dp}}{{dt}} = {t^2}p - p + {t^2} - 1\)

Short answer

The solution is\(p = K{e^{{t^3}/3 - t}} - 1.{\rm{ }}\)

Step-by-step solution

Definition

A differential equation is an equation with one or more derivatives of a function.

Separate variables

Consider the differential equation\(\frac{{dp}}{{dt}} = {t^2}p - p + {t^2} - 1\)

We can rewrite the above equation as

\(\begin{aligned}\frac{{dp}}{{dt}} &= {t^2}p - p + {t^2} - 1\\ &= p\left( {{t^2} - 1} \right) + \left( {{t^2} - 1} \right)\\ &= (p + 1)\left( {{t^2} - 1} \right)\\\frac{{dp}}{{p + 1}} &= \left( {{t^2} - 1} \right)dt\end{aligned}\)

Integrate

Integrating we have:

\(\begin{aligned}\int {\frac{{dp}}{{p + 1}}} &= \int {\left( {{t^2} - 1} \right)} dt + C\\\int {\frac{{dp}}{{p + 1}}} &= \int {{t^2}} dt - \int 1 dt + C\\\ln |p + 1| &= \frac{{{t^{2 + 1}}}}{{2 + 1}} - \frac{{{t^{0 + 1}}}}{{0 + 1}} + C\;\;\;{\rm{ Using formulas }}\int {{x^n}} dx &= \frac{{{x^{n + 1}}}}{{n + 1}} + C,n \ne - 1;\int {\frac{1}{x}} dx\\ &= \ln |x| + Cn|p + 1|\\ &= \frac{1}{3}{t^3} - t + C\end{aligned}\)

Simplify

Solve for\(p\):

\(\begin{aligned}|p + 1| = {e^{{t^3}/3 - t + c}}\\p + 1 = \pm {e^c}{e^{{t^3}/3 - t}}\\p + 1 = K{e^{{t^3}/3 - t}}\\p = K{e^{{t^3}/3 - t}} - 1\end{aligned}\)

Where\(K = \pm {e^c}\), and\(K\) is an arbitrary constant, since\(C\) is an arbitrary constant. Therefore, the solution of differential equation \(\frac{{dp}}{{dt}} = {t^2}p - p + {t^2} - 1\) is\(p = K{e^{{t^3}/3 - t}} - 1.{\rm{ }}\)