Answer to Question #305160 in Differential Equations for She

Question

Find the solution:

2(dS)/(dt) - S/t = 5t^3 S^3

Answer 1

The given equation can be rewritten as, $\dfrac{dS}{dt} - \dfrac{S}{2t} = \dfrac{5}{2}t^3 S^3$ which is a Bernoulli's equation.

Dividing the equation by $S^3$ we get $\dfrac{1}{S^3}\dfrac{dS}{dt} - \dfrac{1}{2S^2t} = \dfrac{5}{2}t^3$

Put $z = \dfrac{1}{S^2}$. Then $\dfrac{dz}{dt} = -\dfrac{2}{S^3}\dfrac{dS}{dt}$ and hence the equation becomes,

$\dfrac{dz}{dt} + \dfrac{2v}{t} = -5t^3$ which is linear in $z$.

The solution is given by,

$\begin{aligned} ze^{\int Pdt} &= \int Q e^{\int Pdt} dt + c\\ ze^{\int \frac{2}{t}dt} &= -\int 5t^3 e^{\int \frac{2}{t} dt} dt + c\quad(P = \frac{2}{t}; Q = -5t^3)\\ zt^2 & = -\int 5t^3 \cdot t^2 dt + c\\ zt^2 & = -\frac{5t^6}{6} + c\\ z &= -\frac{5t^4}{6} + \frac{c}{t^2} = \frac{-5t^6+6c}{6t^2}\\ \frac{1}{S^2} &= \frac{-5t^6+6c}{6t^2}\qquad (\text{Replacing ~} z = \frac{1}{S^2})\\ S^{2} &= \frac{6t^2}{-5t^6+6c}\\ \therefore S& = \pm\sqrt{\frac{6t^2}{-5t^6+6c}} \end{aligned}$