Question

After substituting \[y=e^{-3x}\] and its derivatives into \[\frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}-6y=0\], which expression is obtained on the left-hand side?

पर्याय

• \[-9e^{-3x}-3e^{-3x}-6e^{-3x}=0\]

• \[9e^{-3x}-3e^{-3x}-6e^{-3x}=0\]

• \[9e^{-3x}+3e^{-3x}-6e^{-3x}=0\]

• \[9e^{-3x}-3e^{-3x}+6e^{-3x}=0\]

Answer 1

Here \[\frac{d^{2}y}{dx^{2}}=9e^{-3x}\], \[\frac{dy}{dx}=-3e^{-3x}\], and \[y=e^{-3x}\]. Substitution gives \[9e^{-3x}-3e^{-3x}-6e^{-3x}=0\].