linear differential eqn: (1+t^2)ds+2t(st^2-3(1+t^2)^2)dt=...

T_TEngineer_AdamT_T

New member
Joined
Apr 15, 2007
Messages
24
\(\displaystyle (1+t^2)ds + 2t(st^2 - 3(1+t^2)^2)dt = 0\) when t = 0; t = 2

\(\displaystyle (1+t^2)ds + (2t^3s - 6t(1+t^2)^2)dt\)
i can see that this equation is linear in s

in standard form
\(\displaystyle \frac{ds}{dt} + \frac{2t^3s}{1+t^2} = 6t(1+t^2)\)

the integrating factor is \(\displaystyle e^{(t^2 - \ln{|1+t^2|})} = \frac{e^{t^2}}{1+t^2}\)

\(\displaystyle \frac{e^{t^2}}{1+t^2} \frac{ds}{dt} + \frac{2t^3 e^{t^2}s} {(1+t^2)^2} = 6t(e^{t^2})\)
now i can integrate.... so far is my product rule correct?
 
Top