Question #305160

Find the solution:


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


1
Expert's answer
2022-03-07T17:07:06-0500

The given equation can be rewritten as, dSdtS2t=52t3S3\dfrac{dS}{dt} - \dfrac{S}{2t} = \dfrac{5}{2}t^3 S^3 which is a Bernoulli's equation.


Dividing the equation by S3S^3 we get 1S3dSdt12S2t=52t3\dfrac{1}{S^3}\dfrac{dS}{dt} - \dfrac{1}{2S^2t} = \dfrac{5}{2}t^3


Put z=1S2z = \dfrac{1}{S^2}. Then dzdt=2S3dSdt\dfrac{dz}{dt} = -\dfrac{2}{S^3}\dfrac{dS}{dt} and hence the equation becomes,


dzdt+2vt=5t3\dfrac{dz}{dt} + \dfrac{2v}{t} = -5t^3 which is linear in zz.


The solution is given by,


zePdt=QePdtdt+cze2tdt=5t3e2tdtdt+c(P=2t;Q=5t3)zt2=5t3t2dt+czt2=5t66+cz=5t46+ct2=5t6+6c6t21S2=5t6+6c6t2(Replacing  z=1S2)S2=6t25t6+6cS=±6t25t6+6c\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}


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