The correct option is C y3+2y2−35y1=0
Differentiating the given equation successively, we get
y1=5be5x−7ce−7x⋯(1)
y2=25be5x+49ce−7x⋯(2)
y3=125be5x−343ce−7x⋯(3)
Multiplying equation (1) by 7 and then adding to equation (2), we get
y2+7y1=60be5x⋯(4)
Multiplying equation (1) by 5 and then subtracting it from equation (2), we get
y2−5y1=84ce−7x⋯(5)
Putting the values of b and c, obtained from equation (4) and (5), respectively in equation (3), we get
y3+2y2−35y1=0