The given equation can be rewritten as, dtdS−2tS=25t3S3 which is a Bernoulli's equation.
Dividing the equation by S3 we get S31dtdS−2S2t1=25t3
Put z=S21. Then dtdz=−S32dtdS and hence the equation becomes,
dtdz+t2v=−5t3 which is linear in z.
The solution is given by,
ze∫Pdtze∫t2dtzt2zt2zS21S2∴S=∫Qe∫Pdtdt+c=−∫5t3e∫t2dtdt+c(P=t2;Q=−5t3)=−∫5t3⋅t2dt+c=−65t6+c=−65t4+t2c=6t2−5t6+6c=6t2−5t6+6c(Replacing z=S21)=−5t6+6c6t2=±−5t6+6c6t2
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