Questions: Derive following: yn(t)=c1 e^-3 t+c2 t e^-3 t

Solution Steps

To derive the given function \( y_{n}(t) = c_{1} e^{-3 t} + c_{2} t e^{-3 t} \), we need to find its first and second derivatives with respect to \( t \).

Given the function: \[ y_{n}(t) = c_{1} e^{-3 t} + c_{2} t e^{-3 t} \]

To find the first derivative \( y_{n}'(t) \), we apply the product rule and the chain rule: \[ y_{n}'(t) = \frac{d}{dt} \left( c_{1} e^{-3 t} + c_{2} t e^{-3 t} \right) \] \[ y_{n}'(t) = -3 c_{1} e^{-3 t} + c_{2} \left( \frac{d}{dt} (t e^{-3 t}) \right) \] \[ \frac{d}{dt} (t e^{-3 t}) = t \frac{d}{dt} (e^{-3 t}) + e^{-3 t} \frac{d}{dt} (t) \] \[ = t (-3 e^{-3 t}) + e^{-3 t} = -3 t e^{-3 t} + e^{-3 t} \] Thus, \[ y_{n}'(t) = -3 c_{1} e^{-3 t} + c_{2} (-3 t e^{-3 t} + e^{-3 t}) \] \[ y_{n}'(t) = -3 c_{1} e^{-3 t} - 3 c_{2} t e^{-3 t} + c_{2} e^{-3 t} \]

To find the second derivative \( y_{n}''(t) \), we differentiate \( y_{n}'(t) \): \[ y_{n}''(t) = \frac{d}{dt} \left( -3 c_{1} e^{-3 t} - 3 c_{2} t e^{-3 t} + c_{2} e^{-3 t} \right) \] \[ y_{n}''(t) = -3 c_{1} \frac{d}{dt} (e^{-3 t}) - 3 c_{2} \frac{d}{dt} (t e^{-3 t}) + c_{2} \frac{d}{dt} (e^{-3 t}) \] \[ = -3 c_{1} (-3 e^{-3 t}) - 3 c_{2} (-3 t e^{-3 t} + e^{-3 t}) + c_{2} (-3 e^{-3 t}) \] \[ = 9 c_{1} e^{-3 t} + 9 c_{2} t e^{-3 t} - 3 c_{2} e^{-3 t} - 3 c_{2} e^{-3 t} \] \[ = 9 c_{1} e^{-3 t} + 9 c_{2} t e^{-3 t} - 6 c_{2} e^{-3 t} \]

Final Answer