Q. $y = 2e^{2x} - e^{- x}$ is a solution of the differential equation
A) $y_{2} + y_{1} + 2y = 0$
B) $y_{2} - y_{1} + 2y = 0$
C) $y_{2} + y_{1} = 0$
D) $y_{2} - y_{1} - 2y = 0$
Explanation
(d)
Given, $y = 2e^{2x} - e^{- x}$
$\Rightarrow$ $y_{1} = 4e^{2x} + e^{- x}$
$\Rightarrow$ $y_{2} = 8e^{2x} - e^{- x}$
$\Rightarrow$ $y_{2} = 4e^{2x} + e^{- x} + 4e^{2x} - 2e^{- x}$
$\Rightarrow$ $y_{2} = y_{1} + 2(2e^{2x} - e^{- x})$
$\Rightarrow$ $y_{2} = y_{1} + 2y$
$\Rightarrow$ $y_{2} = y_{1} + 2y$
$\Rightarrow$ $y_{2} - y_{1} - 2y = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.