00:01
Hi, so to solve this question, we'll first consider the auxiliary polynomial i .e.
00:07
P of d equals to r -1 the whole square.
00:13
The auxiliary equation is r -1 the whole square equals to 0.
00:20
R1 is equal to 1.
00:23
The roots of the auxiliary equations are real.
00:26
Therefore, according to the theorem from we get the roots as e raised to rx, x e raised to rx.
00:42
Y1 of x is equal to e raised to x.
00:48
Y2 of x is equal to x e raised to x.
00:52
So y1 of x, y2 of x is equal to e raised to x, x e raised to x.
01:04
Yc of x will be equal to c1 e raised to x plus c2 x e raised to x.
01:13
Now we have f of x is equal to 6 e raised to x.
01:20
F is in the form of x raised to k e raised to ax.
01:32
So further this can be then written as a of d equals to d minus a raised to k plus 1...