Finding an initial value in an ODE [closed]

Given an ODE, $y'(t) = f(t, y), t \in [a, b], y(a) = \alpha$, and a value I, find an $\alpha \in R$ so that the solution satisfies $\int_{a}^{b}y(t)dt = I$.

Can someone explain to me what techniques I should use. You needn't go into too much detail.

Solution 1:

Assuming from the tags that this is a numerics problem, you can view $\alpha \mapsto \int_a^b y(t) dt-I$ as an expensive-to-evaluate black box function (which you could implement numerically by using an ODE solver together with a numerical integrator), and then use scalar root finders on it.

A more elegant way to do it is to look at $Y(t)=\int_a^t y(s) ds$. Notice that regardless of the value of $\alpha$ you have $Y''=f(t,Y'),Y(a)=0$, and that if $\alpha$ is selected correctly then $Y(b)=I$. So you can solve that second order BVP using any of various solvers for second order BVPs, and then extract $Y'(0)$ from your representation of the solution. The method of the previous paragraph is equivalent to the shooting method for this BVP.