We can now solve 1 + u = ±Ket2/2 + 6t for u. u = −1 ± Correct: Your answer is correct. , where K > 0 Notice that u = −1 is a solution. This means that K could also be 0. Since we now know that K ≥ 0 we can let A represent any arbitrary constant and replace ±K with +A to create a more simplified equation. So we could simply write