SOLVING LINEAR EQUATIONS
๐ฆ(4)โ6๐ฆโฒโฒโฒ+7๐ฆโฒโฒ+6๐ฆโฒโ8๐ฆ=0
๐ฆ(0)=โ2,๐ฆโฒ(0)=โ8,๐ฆโฒโฒ(0)=โ14,๐ฆโฒโฒโฒ(0)=โ62
The characteristic polynomial is:
๐4โ6๐3+7๐2+6๐โ8=0
The constant term ir ๐๐=โ๐. From the rational root theorem, the possible roots are
ยฑ1,ยฑ2,ยฑ4, ยฑ8.
Then ๐1(1)=1โ6+7+6โ8=0 so by the remainder theorem, ๐=1 is a lot. Dividing
๐1(1) by ๐โ1 using synthetic division, we obtain:
1
-6
7
6
-8
1
1
-5
2
8
1
-5
2
8
0
Let ๐2(๐)=๐3โ5๐2+2๐+8. checking for ๐=1,๐2(1)=1โ5+2+8โ 0, so
we now discard ๐=1,
Moving on to the next wot, ๐=โ1, we obtain
1
-5
2
-8
-1
-1
6
8
-1
-6
8
0
Let ๐3(๐)=๐2โ6๐+8=0. Checking for ๐=โ1,๐3(โ1)=1โ6+8+0.50 we finally
discard ๐=โ1. Since the function is quadratic, it will be inefficient to use the Rational
root. By factoring, we obtain ๐4(๐)=(๐โ4)(๐โ2) Equating both factors to zero, we get
the last two roots ๐=2 and ๐=4.
Therefore, the roots are: โ1,1,2,4. Hence, the general solution is given by:
๐ฆ=๐1๐โ๐ฅ+๐2๐๐ฅ+๐3๐2๐ฅ+๐4๐4๐ฅ
The first, second and third derivatives of ๐ฆ are:
