Question

if cos inverse (y/b) = log (x/n)" prove that
x²(y base n+2) + (2n + 1)x(y base n+1) + 2n² (y base n) = 0​

Answer 1

i think it helps u.... thank you

Answer 2

Answer:

The result is proved.

Step-by-step explanation:

We need to prove that x²yₙ₊₂ + (2n - 1) xyₙ₊₁ + 2n²yₙ = 0

Given that cos⁻¹ ( $\frac{y}{b}$ ) = log ( $\frac{x}{n}$ )ⁿ

=>  $\frac{y}{b}$  = cos [log ( $\frac{x}{n}$ )ⁿ ]

=> y = b cos [log ( $\frac{x}{n}$ )ⁿ ]

We know that log mⁿ = n log m

=> y = b cos [ n log( $\frac{x}{n}$ )]

First of all, we will differentiate the above equation with respect to x

=> $\frac{dy}{dx}$ = b *(-)sin [ n log( $\frac{x}{n}$ )] * n * $\frac{1}{\frac{x}{n} }$ * $\frac{1}{n}$

=> y₁ = -b sin [ n log( $\frac{x}{n}$ )] * $\frac{n}{x}$

=> xy₁ = - nb sin [ n log( $\frac{x}{n}$ )]

We differentiate again, with respect to x.

=> xy₂ + y₁ = -nb [ cos [ n log( $\frac{x}{n}$ )] * n * $\frac{1}{\frac{x}{n} }$ * $\frac{1}{n}$]

=> xy₂ + y₁ = -nb [ cos [ n log( $\frac{x}{n}$ )] * $\frac{n}{x}$

=> x²y₂ + xy₁ = -n²b [ cos [ n log( $\frac{x}{n}$ )]

As, y = b cos [ n log( $\frac{x}{n}$ )],

Therefore, on substituting the value, we get

x²y₂ + x y₁ = -n²y

Now, according to Leibnitz's Theorem,

Dⁿ(uv) = (Dⁿu)v + ⁿC₁ Dⁿ⁻¹u . Dv +  ⁿC₂ Dⁿ⁻²u . D²v + .... + u.Dⁿv

Therefore, differentiating the given equation 'n' times, we get,

Dⁿ(x²y₂) + Dⁿ(x y₁) = Dⁿ(-n²y)

Applying Leibnitz's Theorem,

=>Dⁿ(y₂)  x² + ⁿC₁ Dⁿ⁻¹(y₂).D(x²) + ⁿC₂ Dⁿ⁻²(y₂) . D²(x²) +

Dⁿ(y₁)  x + ⁿC₁ Dⁿ⁻¹(y₁).D(x) = Dⁿ(y) (-n²)

=> yₙ₊₂ . x² + n. yₙ₋₁₊₂.2x + $\frac{n(n-1)}{2}$ .  yₙ₋₂₊₂.2 + yₙ₊₁ . x + n. yₙ₋₁₊₁.1 = -n².yₙ

=> yₙ₊₂ . x² + n. yₙ₊₁.2x + n(n-1)  yₙ. + yₙ₊₁ . x + n. yₙ = -n².yₙ

=> yₙ₊₂ . x² + n. yₙ₊₁.2x + n² yₙ. - n yₙ. + yₙ₊₁ . x + n. yₙ = -n².yₙ

=> yₙ₊₂ . x² + n. yₙ₊₁.2x + yₙ₊₁ . x + n² yₙ  + n².yₙ = 0

=> x²yₙ₊₂ + xyₙ₊₁ (2n+1) + 2n² yₙ = 0

Which is the desired result.

Hence proved.