More Differentiation Calculus?

Question

Show that y = sinθ/θ is a solution of θ(d²y/dθ²) + 2(dy/dθ) + θy = 0

10pts given to first correct answer with clear working, thanks :)

Answer 1

y = sinθ/θ

dy/dθ

= (θcosθ - sinθ)/θ²

(Using the quotient rule)

d²y/dθ²

= [θ²(-θsinθ + cosθ - cosθ) - 2θ(θcosθ - sinθ)]/θ^4

= [-θ²sinθ - 2θcosθ + 2sinθ)]/θ³

(Using the quotient rule again)

Now substitute these values back into the differential equation.

θ(d²y/dθ²) + 2(dy/dθ) + θy

= [-θ²sinθ - 2θcosθ + 2sinθ)]/θ² + 2(θcosθ - sinθ)/θ² + sinθ

= -sinθ + (-2θcosθ + 2sinθ + 2θcosθ - 2sinθ)/θ² + sinθ

= -sinθ + sinθ + 0

= 0

y = sinθ/θ

is a solution of the given differential equation.