Question

The differential equation whose C.F. is \(y=c_1\cos 2x+c_2\sin 2x+c_3e^{-x}\) is

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

Hint:

From \(\cos bx,\sin bx\), write roots \(\pm ib\). From \(e^{ax}\), write root \(a\).

Answer 1

The Correct Option is B

Concept:
The complementary function tells us the roots of the auxiliary equation. If the complementary function contains \[ \cos bx,\ \sin bx \] then the roots are \[ \pm ib \] If it contains \[ e^{ax} \] then the root is \[ a \]

Step 1: Identify roots from the given C.F. Given, \[ y=c_1\cos2x+c_2\sin2x+c_3e^{-x} \] From \[ \cos2x,\ \sin2x \] we get roots \[ m=\pm 2i \] From \[ e^{-x} \] we get root \[ m=-1 \]

Step 2: Form the auxiliary equation. The factors are \[ (m-2i)(m+2i)(m+1)=0 \] Now, \[ (m-2i)(m+2i)=m^2+4 \] So, \[ (m^2+4)(m+1)=0 \]

Step 3: Expand. \[ (m^2+4)(m+1)=m^3+m^2+4m+4 \] Thus, \[ m^3+m^2+4m+4=0 \]

Step 4: Convert to differential equation. Replace \(m\) by \(D\): \[ D^3+D^2+4D+4=0 \] Therefore, \[ \frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}+4\frac{dy}{dx}+4y=0 \]

Step 5: Final answer. \[ \boxed{\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}+4\frac{dy}{dx}+4y=0} \]