00:01
In this question, for the given differential equation, we are asked to check that for k equals 1, the function c1, e to the d plus c2, e to the negative t is a solution of the differential equation by direct substitution.
00:18
So to check that this function is a solution for k equals 1, first we need to write down the differential equation for k equals 1.
00:30
In the case of k equals 1, the differential equation is y double prime of t minus y of t equals to 0.
00:40
Now we need to plug in y and y double prime in the equation.
00:48
However, we first need to calculate y double prime.
00:52
To calculate y double prime, we first need to find y prime of t.
00:57
Y prime of t equals to the derivative of c1, e to the t plus c2, e to the negative t.
01:07
And that's going to be c1 e to the t minus c2 e to the negative t.
01:15
Now we're going to calculate y double prime...