Question

The solution of the differential equation \(\displaystyle \frac{d^3y}{dx^3}+3\frac{d^2y}{dx^2}+2\frac{dy}{dx}=0\) is

• A. \(y=a+be^{-x}+ce^{-2x}\)
• B. \(y=a+be^{x}+ce^{2x}\)
• C. \(y=ae^{-x}+be^{-2x}+ce^{x}\)
• D. \(y=a+be^{-2x}+ce^{-3x}\)

Hint:

For a linear differential equation with constant coefficients, form the auxiliary equation and solve for roots.

Answer 1

The Correct Option is A

We are given: \[ \frac{d^3y}{dx^3}+3\frac{d^2y}{dx^2}+2\frac{dy}{dx}=0. \] Let \[ D=\frac{d}{dx}. \] Then the differential equation becomes: \[ (D^3+3D^2+2D)y=0. \] Take \(D\) common: \[ D(D^2+3D+2)y=0. \] Now factorize: \[ D(D+1)(D+2)y=0. \] Therefore, the auxiliary equation is: \[ m(m+1)(m+2)=0. \] So, \[ m=0,\quad m=-1,\quad m=-2. \] For \(m=0\), the solution is: \[ a. \] For \(m=-1\), the solution is: \[ be^{-x}. \] For \(m=-2\), the solution is: \[ ce^{-2x}. \] Therefore, the complete solution is: \[ y=a+be^{-x}+ce^{-2x}. \] Hence, the correct answer is: \[ y=a+be^{-x}+ce^{-2x}. \]