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?
\(\displaystyle (1+t^2)ds + (2t^3s - 6t(1+t^2)^2)dt\)
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?