\frac{d^3x}{dt^3} – 3\frac{d^2x}{dt^2} – 2\frac{dx}{dt} + 6x = 0 4.

. \frac{d^3x}{dt^3} – 3\frac{d^2x}{dt^2} – 2\frac{dx}{dt} + 6x = 0 4. \frac{d^3x}{dt^3} – 2\frac{d^2x}{dt^2} – 10\frac{dx}{dt} + 8x = 0

The Correct Answer and Explanation is:

Let’s solve the two third-order linear homogeneous differential equations using the characteristic equation method and then provide a clear explanation.

Problem 3

d3xdt3−3d2xdt2−2dxdt+6x=0\frac{d^3x}{dt^3} – 3\frac{d^2x}{dt^2} – 2\frac{dx}{dt} + 6x = 0dt3d3x​−3dt2d2x​−2dtdx​+6x=0

Step 1: Form the characteristic equation:r3−3r2−2r+6=0r^3 – 3r^2 – 2r + 6 = 0r3−3r2−2r+6=0

Try rational roots. r=1r = 1r=1 and r=−1r = -1r=−1 do not work. Try r=3r = 3r=3:33−3(3)2−2(3)+6=27−27−6+6=03^3 – 3(3)^2 – 2(3) + 6 = 27 – 27 – 6 + 6 = 033−3(3)2−2(3)+6=27−27−6+6=0

So r=3r = 3r=3 is a root. Divide the polynomial by (r−3)(r – 3)(r−3):r3−3r2−2r+6=(r−3)(r2−2)r^3 – 3r^2 – 2r + 6 = (r – 3)(r^2 – 2)r3−3r2−2r+6=(r−3)(r2−2)

Solve r2−2=0⇒r=±2r^2 – 2 = 0 \Rightarrow r = \pm \sqrt{2}r2−2=0⇒r=±2​

Solution:x(t)=C1e3t+C2e2t+C3e−2tx(t) = C_1 e^{3t} + C_2 e^{\sqrt{2}t} + C_3 e^{-\sqrt{2}t}x(t)=C1​e3t+C2​e2​t+C3​e−2​t

Problem 4

d3xdt3−2d2xdt2−10dxdt+8x=0\frac{d^3x}{dt^3} – 2\frac{d^2x}{dt^2} – 10\frac{dx}{dt} + 8x = 0dt3d3x​−2dt2d2x​−10dtdx​+8x=0

Step 1: Characteristic equation:r3−2r2−10r+8=0r^3 – 2r^2 – 10r + 8 = 0r3−2r2−10r+8=0

Try r=4r = 4r=4:64−32−40+8=0⇒r=4 is a root64 – 32 – 40 + 8 = 0 \Rightarrow r = 4 \text{ is a root}64−32−40+8=0⇒r=4 is a root

Divide:r3−2r2−10r+8=(r−4)(r2+2r−2)r^3 – 2r^2 – 10r + 8 = (r – 4)(r^2 + 2r – 2)r3−2r2−10r+8=(r−4)(r2+2r−2)

Solve quadratic:r=−2±(2)2+82=−2±122=−1±3r = \frac{-2 \pm \sqrt{(2)^2 + 8}}{2} = \frac{-2 \pm \sqrt{12}}{2} = -1 \pm \sqrt{3}r=2−2±(2)2+8​​=2−2±12​​=−1±3​

Solution:x(t)=C1e4t+C2e(−1+3)t+C3e(−1−3)tx(t) = C_1 e^{4t} + C_2 e^{(-1 + \sqrt{3})t} + C_3 e^{(-1 – \sqrt{3})t}x(t)=C1​e4t+C2​e(−1+3​)t+C3​e(−1−3​)t

Explanation

To solve higher-order linear differential equations with constant coefficients, we use the characteristic equation method. This technique transforms the differential equation into a polynomial by replacing derivatives with powers of a variable rrr. For example, d3xdt3\frac{d^3x}{dt^3}dt3d3x​ becomes r3r^3r3, d2xdt2\frac{d^2x}{dt^2}dt2d2x​ becomes r2r^2r2, and so on. Solving the resulting polynomial yields the roots, which guide the form of the solution.

For Problem 3, we obtained the characteristic equation r3−3r2−2r+6=0r^3 – 3r^2 – 2r + 6 = 0r3−3r2−2r+6=0. Using factorization, we found the roots r=3,2,−2r = 3, \sqrt{2}, -\sqrt{2}r=3,2​,−2​. These real roots produce exponential solutions, leading to the general solution x(t)=C1e3t+C2e2t+C3e−2tx(t) = C_1 e^{3t} + C_2 e^{\sqrt{2}t} + C_3 e^{-\sqrt{2}t}x(t)=C1​e3t+C2​e2​t+C3​e−2​t.

In Problem 4, the characteristic equation is r3−2r2−10r+8=0r^3 – 2r^2 – 10r + 8 = 0r3−2r2−10r+8=0. We found r=4r = 4r=4 and factored the remaining quadratic to get complex roots r=−1±3r = -1 \pm \sqrt{3}r=−1±3​. These also lead to exponential solutions, resulting in the general solution x(t)=C1e4t+C2e(−1+3)t+C3e(−1−3)tx(t) = C_1 e^{4t} + C_2 e^{(-1+\sqrt{3})t} + C_3 e^{(-1-\sqrt{3})t}x(t)=C1​e4t+C2​e(−1+3​)t+C3​e(−1−3​)t.

These solutions describe how the system evolves over time. The nature of the roots—real or complex—determines whether the behavior is exponential growth/decay or oscillatory.