Question

y=(e^3x)(sin4x)

d2y/dx2+ a dy/dx+by=0

find the value of a-b

Answer 1

dy/dx = d((e^3x)(sin4x))/dx = d(e^3x)/dx *sin4x + e^3x * d(sin4x)/dx
= 3*e^3x *sin4x + 4*e^3x *cos4x
Again ,
d^2y/dx^2 = d(dy/dx)/dx
= 3*(3e^3x*sin4x +4*e^3x*cos4x) + 4*(3*e^3x*cos4x - 4*e^3x *sin4x)
= - 7e^3x *sin4x + 24 e^3x * cos4x

Answer 2

So,
d^2y/dx^2+ a dy/dx+by=0
=> - 7e^3x *sin4x + 24 e^3x * cos4x + a(3*e^3x *sin4x + 4*e^3x *cos4x) + b(e^3x)*sin4x =0
=> (-7+3a+b)*e^3x *sin4x + (24+4a) *e^3x *cos4x =0

Answer 3

So, -7+3a+b=0 and 24+4a=0
==> a= -6 and b=25
and a-b= -6-25=-31