00:01
In this question we are considering the differential equation y double prime plus 2y prime minus 3y equal negative 16 x exponential x.
00:14
The auxiliary equation is r squared plus 2r minus 3 equal 0.
00:27
It has the solution r equal to 1 or r equal to negative 3.
00:32
So if we define this to be t y, then we view this as an operator operating on y, then we have exponential t exponential x and the t exponential negative 3x they are zero because these two are the corresponding solution, linearly independent solution of the complementary equation.
01:23
In order to make the right -hand side to have x exponential x, we try if we try exponential x, then we get zero.
01:33
And if we try x exponential x, what we will get is something times exponential x.
01:43
Because this is what we use to solve the differential equation if the right hand side is exponential x.
01:52
So we need to try a particular solution of the form.
02:04
A .x squared exponential x plus bx exponential x.
02:18
So we write it as a x square plus bx exponential x.
02:25
Exponential x.
02:32
Then we have yp prime.
02:38
This is 2ax plus b exponential x plus a x plus bx exponential x...