I need to solve this set of differential equations {y'+(t/2)1, y(0)=1} and then find the limiting value as t approaches infinity. I'm not sure where to start. I found the answer to the differential equation to be: y(t)=e^-t^2/4[e^(r^2/4)dr +Ce^(-t^2/4)where the [ sign represents the definite integral from 0 to t. Now I'm not sure how to evaluate the limit of this as t approahes 0. Any help would be appreciated!
To solve the given set of differential equations, we can use the method of integrating factors. Let's go through the steps:
1. Start with the given differential equation: y' + (t/2) = 1.
2. Rearrange the equation in the standard form: y' = 1 - (t/2).
3. Now, integrate both sides of the equation with respect to t to find the general solution:
∫y' dt = ∫(1 - t/2) dt.
This gives: y = t - (t^2/4) + C, where C is the constant of integration.
4. We are given the initial condition y(0) = 1. We can substitute this into the general solution:
1 = 0 - (0^2/4) + C.
Simplifying, we find: C = 1.
5. Substituting this value of C back into the general solution, we get the particular solution:
y = t - (t^2/4) + 1.
Now, to find the limiting value of y as t approaches infinity, we consider the behavior of the quadratic term. The term -(t^2/4) approaches negative infinity as t approaches infinity, which means it dominates the linear term (t) and the constant term (1) in the expression for y.
Hence, the limiting value of y as t approaches infinity is negative infinity: lim (y) = -∞.