00:01
In this problem, we wish to find the solution to the given initial value problem, y prime equals x to negative second minus x to negative third for y negative 1 equals 0, and we want to determine then on what interval is the solution valid.
00:13
This question is challenging our understanding of how to apply definite or rather indefinite integration towards solving initial value problems and differential equations.
00:23
Rather to solve initial value problems using the indefinite integral, we use the indefinite integral formula integral fxdx equals capital fx plus c, where c is a constant of integration, we solve, or given the initial value.
00:35
To utilize this form, what we have to do is to separate dy into dydx, and then multiply both sides.
00:40
Thus, we have integral dy equals integral x negative second minus x negative 30x, which gives y equals negative 1 over x plus 1 over 2x squared plus constant integration c...