Question

Complementary function of the differential equation $(D^3 - 8)y = x$ is

• A. $c_1e^{2x} + e^x\{c_2\cos(x\sqrt{3}) + c_3\sin(x\sqrt{3})\}$
• B. $c_1e^{2x} + e^{-x}(\cos\sqrt{3} + \sin\sqrt{3})$
• C. $c_1e^{-2x} + c_2e^{-x} + c_3\cos x$
• D. $c_1e^{2x} + e^{-x}\{c_2\cos(x\sqrt{3}) + c_3\sin(x\sqrt{3})\}$

Hint:

Complex roots always appear in conjugate pairs. If you find one complex root, the other is guaranteed to be its conjugate. The real part goes into the exponential factor, and the imaginary part goes inside the sine and cosine.

Answer 1

The Correct Option is D

Step 1: Form the Auxiliary Equation: Replace $D$ with $m$ and set the equation to zero: $$m^3 - 8 = 0$$

Step 2: Solve for Roots: Using the identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$: $$(m - 2)(m^2 + 2m + 4) = 0$$

• Root 1: $m - 2 = 0 \implies m_1 = 2$

• Roots 2 & 3: Solve $m^2 + 2m + 4 = 0$ using the quadratic formula: $$m = \frac{-2 \pm \sqrt{2^2 - 4(1)(4)}}{2(1)} = \frac{-2 \pm \sqrt{-12}}{2} = \frac{-2 \pm 2i\sqrt{3}}{2} = -1 \pm i\sqrt{3}$$

Step 3: Write the CF:
• For the real root ($m=2$): $c_1e^{2x}$

• For the complex roots ($\alpha \pm i\beta = -1 \pm i\sqrt{3}$): $e^{\alpha x} (c_2 \cos \beta x + c_3 \sin \beta x)$ $$\text{Result} = e^{-x} (c_2 \cos(x\sqrt{3}) + c_3 \sin(x\sqrt{3}))$$
Combining them: $$CF = c_1e^{2x} + e^{-x}\{c_2\cos(x\sqrt{3}) + c_3\sin(x\sqrt{3})\}$$