Asked by George
solve the initial value problem and find the interval of existence...
cos(y) dy/dt = -t sin(y)/ (1+t^2) , y(1)=pi/2
cos(y) dy/dt = -t sin(y)/ (1+t^2) , y(1)=pi/2
Answers
Answered by
Steve
cos(y) dy/dt = -t sin(y)/ (1+t^2)
cosy/siny dy = -t/(1+t^2) dt
ln(siny) = -1/2 ln(1+t^2) + c
sint = c/√(1+t^2)
sin(π/2) = c/√1
c = 1
y = arcsin(1/√(1+t^2))
cosy/siny dy = -t/(1+t^2) dt
ln(siny) = -1/2 ln(1+t^2) + c
sint = c/√(1+t^2)
sin(π/2) = c/√1
c = 1
y = arcsin(1/√(1+t^2))