What is the solution for (dp)/(dt)=t^2p-3p+t^2-3 ?

The solution for (dp)/(dt)=t^2p-3p+t^2-3 is p=e^{(t^3)/3-3t+c_{1}}-1

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(dp)/(dt)=t^2p-3p+t^2-3

\frac{d p}{d t} = t^{2} p - 3 p + t^{2} - 3

p = e^{\frac{t ^{3}}{3} - 3 t + c_{1}} - 1

Solution steps

\frac{d p}{d t} = t^{2} p - 3 p + t^{2} - 3

Solve separable ODE:

p = e^{\frac{t ^{3}}{3} - 3 t + c_{1}} - 1

p = e^{\frac{t ^{3}}{3} - 3 t + c_{1}} - 1

\frac{( t ^{2} - 12 t + 11 ) d y}{d t} = y

\frac{d x}{d t} = 3 x

\frac{d y}{d x} = \frac{x ^{2} + 1}{y}

\frac{d y}{d x} = - \frac{x}{y e ^{x^{2}}}, y (0) = 1

\frac{d x}{d t} = 2 t x^{2}

What is the solution for (dp)/(dt)=t^2p-3p+t^2-3 ?
The solution for (dp)/(dt)=t^2p-3p+t^2-3 is p=e^{(t^3)/3-3t+c_{1}}-1