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)
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))
To solve the initial value problem, we have the differential equation:
cos(y) dy/dt = (-t sin(y))/(1+t^2)
First, we can separate the variables by multiplying both sides of the equation by dt/cos(y):
dy/cos(y) = (-t sin(y))/(1+t^2) dt
Next, we can integrate both sides of the equation. Integrating dy/cos(y) is a bit tricky, but we can use a trigonometric identity to simplify it:
∫dy/cos(y) = ∫(-t sin(y))/(1+t^2) dt
Let's use the identity cos^2(y) = 1 - sin^2(y) to rewrite the numerator in terms of cos(y):
∫dy/cos(y) = ∫(-t cos(y) sin(y))/(1+t^2) dt
Now, let's substitute u = 1+t^2 and du = 2t dt:
∫dy/cos(y) = ∫(-1/2) (cos(y) sin(y))/u du
Using a substitution v = cos(y), we can rewrite the integral as:
∫(-1/2) (sin(y))/(v^2) dv
This integral can be easily evaluated:
∫(-1/2) (sin(y))/(v^2) dv = (1/2v) + C
Re-substituting v = cos(y) and u = 1 + t^2:
(1/2v) + C = (1/2cos(y)) + C = (1/2√(u-1)) + C
Now, let's solve for y:
(1/2√(u-1)) + C = y
Substituting back u = 1 + t^2:
(1/2√(1 + t^2 - 1)) + C = y
(1/2√(t^2)) + C = y
(1/2|t|) + C = y
Now, let's include the initial condition y(1) = π/2 :
(1/2|1|) + C = π/2
1/2 + C = π/2
C = π/2 - 1/2
So our solution is:
(1/2|t|) + C = y
(1/2|t|) + π/2 - 1/2 = y
Finally, let's find the interval of existence. Since cos(y) is in the denominator of the original differential equation, we need to avoid values of y where cos(y) equals zero. This means we need to exclude y = (2n+1)π/2 for all integers n, since these are the values where cos(y) equals zero.
Therefore, the interval of existence is (-∞, ∞) excluding the values y = (2n+1)π/2 for all integers n.