Question

then
show
that
3) [July 2014 July 2017]: If y= 3 cos (log x) + 4 sin (log x),
x y2 + xy + y=0
Solution​

Answer 1

Answer:

Given, \bf{y=3cos(logx)+4sin(logx)}y=3cos(logx)+4sin(logx)

we have to proof that \bf{x^2y_2+xy_1+y}=0x

2

y

2

+xy

1

+y=0

y = 3cos(logx) + 4sin(logx) ------(1)

differentiate y with respect to x,

dy/dx = y₁ = 3. d(coslogx)/dx + 4. d(sinlogx)/dx

= 3{-sin(logx)}1/x + 4{cos(logx)}1/x

= {-3sin(logx) + 4cos(logx)}/x

xy₁ = {-3sin(logx) + 4cos(logx)}

again differentiate with respect to x ,

d(xy₁ )/dx = -3. d{sin(logx)}/dx + 4. d{cos(logx)}/dx}

=> x. dy₁ /dx + y₁ .dx/dx = -3.cos(logx)1/x - 4sin(logx).1/x

=> xy₂ + y₁ = -[3cos(logx) + 4sin(logx)]/x

=> x²y₂ + xy₁ = -[3cos(logx) + 4sin(logx)]

=> x²y₂ + xy₁ = -y [ from equation (1),

=> x²y₂ + xy₁ + y = 0 \textbf{\underline{hence proved}}

hence proved