The second-order homogeneous linear differential equation whose characteristic equation has roots
−
3
8
+
1
4
𝑖
and
−
3
8
−
1
4
𝑖
is:

Hint: A quadratic equation whose roots are
𝛼
and
𝛽
is:
𝑋
2
−
(
𝛼
+
𝛽
)
𝑋
+
𝛼
𝛽
=
0
.

Question

The second-order homogeneous linear differential equation whose characteristic equation has roots 
−
3
8
+
1
4
𝑖
 and 
−
3
8
−
1
4
𝑖
 is:

Hint: A quadratic equation whose roots are 
𝛼
 and 
𝛽
 is:   
𝑋
2
−
(
𝛼
+
𝛽
)
𝑋
+
𝛼
𝛽
=
0
.

BlackTom AI · Accepted Answer

The problem asks for the second-order homogeneous linear differential equation whose characteristic roots are -3/8 ± i/4. 
First, recall that if the roots of the characteristic equation are α and β, the corresponding quadratic is X^2 − (α + β) X + αβ = 0. Here, the two roots are complex conjugates: α = -3/8 + i/4 and β = -3/8 − i/4.
Compute the sum: α + β = (-3/8 + i/4) + (-3/8 − i/4) = -3/4. 
Compute the product: αβ = (-3/8 + i/4)(-3/8 − i/4) = (-3/8)^2 + (1/4)^2 = 9/64 + 1/16 = 9/64 + 4/64 = 13/64.
Thus the quadratic with these roots is X^2 − (−3/4) X + 13/64 = X^2 + (3/4) X + 13/64 = 0.
To convert to a standard differential equation form, use the fact that if the characteristic equation is r^2 + p r + q = 0, then the corresponding differential equation is y'' + p y' + q y = 0. Here, p = 3/4 and q = 13/64.
Clear denominators by multiplying through by 64 to obtain integer coefficients: 64 y'' + 48 y' + 13 y = 0.
Therefore, the differential equation that has the given roots is 64 x'' + 48 x' + 13 x = 0. 
A quick consistency check: the derived p and q match the real and imaginary parts of the roots when forming the quadratic, and the multiplication by 64 yields clean integer coefficients as shown, which aligns with typical expectations for such problems.

SMAT011 Weekly Quiz 9 |LA009

The second-order homogeneous linear differential equation whose characteristic equation has roots − 3 8 + 1 4 𝑖 and − 3 8 − 1 4 𝑖 is: Hint: A quadratic equation whose roots are 𝛼 and 𝛽 is: 𝑋 2 − ( 𝛼 + 𝛽 ) 𝑋 + 𝛼 𝛽 = 0 .

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