1.875(4+2y)^(2/3)=2.52y^(5/3)

Question

help please?

Answer 1

Rewrite:

(4+2y)^⅔=(2.52/1.875)y^5/3, cube both sides:

(4+2y)²=(2.52/1.875)³y⁵; let a=(2.52/1.875)³=0.60693 approx.

4+4y+y²=ay⁵; ay⁵-y²-4y-4=0.

Let f(y)=ay⁵-y²-4y-4, f'(y)=5ay⁴-2y-4.

Using Newton's Method, y_n+1=y_n-f(y_n)/f'(y_n), let y₀=2.

y₁=1.9156..., y₂=1.9058..., y₃=1.9056..., y₄=1.905686128..., y₅=1.905686128457...

So the approximate solution is y=1.905686128457.