00:01
In this problem, we've been given a differential equation here and some y, and we want to check that this y of x is a solution to our differential equation.
00:12
So to do that, first we check that y is at least twice differentiable since we have y double prime here.
00:18
Well, it is because it's a function of addition and e to the x, which is going to be infinitely differentiable.
00:27
So we can go ahead and take these derivatives.
00:29
Y prime of x is equal to c1e to the x plus we're going to do the chain rule here, minus 2, c2, e to the minus 2x.
00:43
And y double prime of x is going to be equal to c1e to the x, plus we're going to do the chain rule again.
00:53
Negative 2 times negative 2 is 4.
00:56
C2 e.
00:57
To the negative to x.
01:00
Now that we have these, we want to plug them into our differential equation and make sure we get zero.
01:06
So we're going to take this y double prime and plug it in here, and the y prime and plug it in here, and y of x plugged in to here.
01:17
So let's go ahead and do that.
01:20
Y double prime is c1e to the x plus 4c2e to the next, and negative 2x plus y prime is going to be c1 e to the x minus 2 c2 e to the minus 2 x and finally minus 2 y is c1 e to the x plus c2 e to the negative 2 x so we take this and we can go ahead and distribute this minus 2...