00:01
For this problem, we are asked to find the solution of the given initial value problem.
00:04
Y times dy by dx equals 2x over the square root of 1 plus y squared.
00:11
So on what we'll do is first multiply both sides by the square root of 1 plus y squared.
00:18
So we have y root 1 plus y squared, y prime on the left hand side, or actually i'll write it this way, y root 1 plus y squared, dy on the left hand side, and 2x dx on the right hand side.
00:30
Then we simply integrate over each side.
00:34
On the left -hand side, the integral of y -route 1 plus y squared is going to be 1 over 3 times 1 plus y squared to the power of 3 over 2.
00:45
And on the right -hand side, the integral of 2x is going to be x squared plus a constant.
00:50
Now, we have that y of 0 equals 0.
00:55
So we have 1 over 3 times 1 plus 0 to the power of 3 over 2 must be equal to 0 plus c.
01:08
So we have c must be equal to 1 over 3 because 1 to the power of 3 over 2 would simply be 1.
01:15
So we have c equals 1 over 3...