00:02
Here we are given this differential equation dy by dt equal to y and two initial conditions are given first is y at 0 equal to 1 and second initial condition is y at 0 equal minus 1.
00:20
So we have to find solution using both of these initial conditions.
00:24
First of all let us find a solution of this differential equation.
00:28
So here we can write it as dy over y equal to dt.
00:36
On integrating both sides we will get ln y is equal to 2t plus c1.
00:47
So this is the general form of the general solution of the differential equation.
00:52
Let us use the first initial condition which is y at 0 is equal to 1.
00:56
So this will be ln 1 equal 2 into 0 plus c1.
01:04
Ln 1 is 0 so that means value of c1 is also equal to 0.
01:10
So we will get the solution equal to ln y equal 2t plus 0.
01:19
So this will be the solution.
01:22
Similarly let us use this initial condition now...