Consider the equation, y=ex(A‌cos‌x+B‌sin‌x) So, y′=ex(−A‌sin‌x+B‌cos‌x)+ex(A‌cos‌x+B‌sin‌x) y′−y=ex(−A‌sin‌x+B‌cos‌x) y"−y′=ex(−A‌cos‌x−B‌sin‌x)+ex(−A‌sin‌x+B‌cos‌x) =−y+y′−y This implies, y"−2y′+2y=0