If d^n y d x^n =y_n and y=e^ x +e^ - x , then 4 x y₂+2 y₁=

Step-by-step solution

y=e^ x +e^ - x y₁=e^ x 1 2 x -e^ - x ( 1 2 x )y₁= 1 2 x (e^ x -e^ - x ) ...(i) y₂= 1 2 x ( e^ x 1 2 x - (-e^ - x ) ( 1 2 x ))+ (e^ x -e^ - x ) 2 . (- 1 2 ) 1 x^ 3 / 2 ...(ii) from (ii) 2 y₁= 1 x (e^ x -e^ - x ) ...(iii) 4 x y₂=e^ x +e^ - x - ( e^ x -e^ - x x ) ...(iv) Adding (iii) & (iv), 2y₁+4 x y₂=e^ x +e^ - x =y