Question

\[ e^{-x}\left(c_1\cos\sqrt{3}\,x+c_2\sin\sqrt{3}\,x\right)+c_3e^{2x} \] is the general solution of

• A. \(\dfrac{d^3y}{dx^3}+4y=0\)
• B. \(\dfrac{d^3y}{dx^3}-8y=0\)
• C. \(\dfrac{d^3y}{dx^3}+8y=0\)
• D. \(\dfrac{d^3y}{dx^3}-2\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}-2y=0\)

Hint:

From \(e^{\alpha x}\cos\beta x\) and \(e^{\alpha x}\sin\beta x\), write roots \(\alpha\pm i\beta\).

Answer 1

The Correct Option is B

Concept:
From the general solution, we can identify the roots of the auxiliary equation. If the solution contains \[ e^{\alpha x}\cos \beta x,\quad e^{\alpha x}\sin \beta x \] then the roots are \[ \alpha\pm i\beta \]

Step 1: Identify roots from the given solution. Given solution is \[ e^{-x}(c_1\cos\sqrt{3}x+c_2\sin\sqrt{3}x)+c_3e^{2x} \] From \[ e^{-x}\cos\sqrt{3}x,\quad e^{-x}\sin\sqrt{3}x \] we get roots \[ m=-1\pm i\sqrt{3} \] From \[ e^{2x} \] we get root \[ m=2 \]

Step 2: Form the auxiliary equation. \[ (m-2)\left[(m+1)^2+(\sqrt{3})^2\right]=0 \] \[ (m-2)\left[(m+1)^2+3\right]=0 \] \[ (m-2)(m^2+2m+1+3)=0 \] \[ (m-2)(m^2+2m+4)=0 \]

Step 3: Expand. \[ (m-2)(m^2+2m+4) \] \[ =m^3+2m^2+4m-2m^2-4m-8 \] \[ =m^3-8 \] Thus the auxiliary equation is \[ m^3-8=0 \]

Step 4: Convert to differential equation. Replace \(m\) by \(D\): \[ D^3-8=0 \] Therefore, \[ \frac{d^3y}{dx^3}-8y=0 \]

Step 5: Final answer. \[ \boxed{\frac{d^3y}{dx^3}-8y=0} \]