00:01
In this problem, we need to solve the initial value problem ivp, y' ' plus 4y is equal to 0 and the initial conditions y at 0 is equal to 5 and y ' at 0 is equal to minus 7.
00:22
First step is to write the auxiliary equation for this differential equation that is m square plus 4 is equal to 0 and m square would be minus 4, m is equal to plus or minus 2i.
00:45
If the roots of this equation are as like alpha plus i beta, if they are the complex numbers, the solution can be given as the solution of this differential equation can be given as e power alpha t into c1 cos 2 cos beta t plus c2 sin beta t.
01:13
This would be the answer, this would be the general solution.
01:17
So, if m is of the form like this, then this is the solution.
01:24
So, here there is no real part, here alpha is equal to 0 and beta is equal to alpha is equal to.
01:35
So, substituting this what do we get? y at t is equal to e power 0 t into c1 cos beta t plus c2 sin beta is here 2t, here beta is 2t right 2.
01:52
So, 2t plus sin 2t.
01:54
So, e power 0 into t is e power 0, e power 0 is 1.
02:00
So, the solution is c1 cos 2t plus c2 sin 2t, this is the general solution.
02:08
How to get the c1 and c2 values from here from the initial conditions? plug in the initial condition y at 0 is equal to 5.
02:19
Using this substitute here t is equal to 0 in this one, from 1 what do we get? y at t, t is equal to 0, y of 0 is equal to y at 0 is equal to c1 cos t, here c1 cos 0 plus c2 sin 0.
02:44
That implies y at 0 is given 5, 5 is equal to c1 cos 0 is 1 whereas sin 0 is 0...