Q2E

In Problems 1–7, convert the given initial value problem into an initial value problem for a system in normal form.

$y'' (t) = \cos (t - y) + y^{2} (t); y (0) = 1, y' (0) = 0$

Verified

Answer

$x'_{1} (t) = x_{2} (t) x'_{2} (t) = \cos (t - x_{1}) + {x^{2}}_{1} (t) x_{1} (0) = 1 x_{2} (0) = 0$

1Step 1: express the equation in form of x

Here given $y'' (t) = \cos (t - y) + y^{2} (t)$

Denote,

$x_{1} (t) = y (t) x_{2} (t) = y' (t)$

The equation transforms as;

$x'_{1} (t) = x_{2} (t) x'_{2} (t) = \cos (t - x_{1}) + {x^{2}}_{1} (t)$

2Step 2: the initial conditions

The given initial conditions are $y (0) = 1, y' (0) = 0 .$

Initial conditions after transformations;

$x_{1} (0) = 1 x_{2} (0) = 0$

This is the required result.