Hi All,
Can somebody please help me solve this IVP? Or simply explain ivp's to me so I know what I'm supposed to do with this?
(1+t^2)dy/dt + 2ty = -t sin(t), y(pi) = 0
Thank you
Okay. Here is what ive done, and now Im stuck on a part.
(1+t^2)dy/dt + 2ty = -t sin(t), y(pi) = 0
dy/dt + (2ty/(1+t^2))*y = (-t sin(t))/(1+t^2)
Then I find the integrating factor, which I think is
=e^(y*ln(1+t^2))
= (1 +t^2)^y
Multiplying this through, you get:
[((1+t^2)^y) * y] ' = [-(((1+t^2)^y)*t*sin(t))/(1+t^2)]
Then I think I'm supposed to integrate both sides. I know the l.h.s. is just
((1+t^2)^y)*y
but im having trouble integrating [-(((1+t^2)^y)*t*sin(t))/(1+t^2)]
Have I done something wrong here? Why does your answer look so neat compared to mine?
I know that once I have that I can apply the initial condition and workout the particular solution.