If #x^2 y=a cos⁡#x, where #a# is a constant, show that #x^2 (d^2 y)/(dx^2 )+4x dy/d +(x^2+2)y=0 #?

1 Answer
Nov 26, 2017

We have:

# x^2 y=a cos⁡x #

If we differentiate implicitly wrt #x# and apply the product rule then:

# (x^2)(d/dx y) + (d/dx x^2)(y) = -asinx #

# :. x^2dy/dx + 2xy = -asinx #

Differentiating again we wgt:

# (x^2)(d/x dy/dx) + (d/dx x^2)(dy/dx) + (2x)(d/dxy) + (d/dx2x)(y) = -acosx #

# :. x^2 (d^2y)/(dx^2) + 2xdy/dx + 2xdy/dx + 2y = -acosx #

# :. x^2 (d^2y)/(dx^2) + 4xdy/dx + 2y + acosx = 0 #

And using the initial equation:

# x^2 (d^2y)/(dx^2) + 4xdy/dx + 2y + x^2 y = 0 #

Leading to:

# x^2 (d^2y)/(dx^2) + 4xdy/dx + (x^2+2) y = 0 \ \ \ \ # QED