Question

Prove and deduce that
y=(sin-1x)2
Then
Prove
1) (1-x2)y2-xy1-2=0
2)(1-x2)yn+2-(2n+1)xyn+1-n2yn=0

Answer 1

$\textbf{Given:}$

$y=(sin^{-1}x)^2$

$\textbf{To prove:}$

$1.(1-x^2)y_2-xy_1-2=0$

$2.(1-x^2)y_{n+2}-(2n+1)xy_{n+1}-n^2Y_n=0$

$\textbf{Solution:}$

$\text{Consider,}$

$y=(sin^{-1}x)^2$

$\text{Differentiate with respect to x}$

$y_1=2(sin^{-1}x)\dfrac{1}{\sqrt{1-x^2}}$

$\text{Squaring on bothsides, we get}$

${y_1}^2=4(sin^{-1}x)^2\dfrac{1}{1-x^2}$

$(1-x^2){y_1}^2=4y$

$\text{Differentiate with respect to x}$

$(1-x^2)2y_1\,y_2+{y_1}^2(-2x)=4y_1$

$\text{Divide bothsides by 2y_1 }$

$(1-x^2)y_2-xy_1=2$

$\implies\boxed{\boxed{\bf(1-x^2)y_2-xy_1-2=0}}$...........(1)

$\text{Differentiate (1) with respect to x}$

$(1-x^2)y_3+y_2(-2x)-(xy_2+y_1)=0$

$\implies\boxed{(1-x^2)y_3-3xy_2-y_1=0}$.........(2)

$\text{Differentiate (2) with respect to x}$

$(1-x^2)y_4+y_3(-2x)-3(xy_3+y_2)-y_2=0$

$\implies\boxed{(1-x^2)y_4-5xy_3-4y_2=0}$.........(3)

$\text{Differentiate (3) with respect to x}$

$(1-x^2)y_5+y_4(-2x)-5(xy_4+y_3)-4y_3=0$

$\implies\boxed{(1-x^2)y_5-7xy_4-9y_3=0}$

$\text{Proceeding like this, we get}$

$\boxed{\boxed{\bf(1-x^2)y_{n+2}-(2n+1)xy_{n+1}-n^2y_n=0}}$

Answer 2

Answer:

(

1

−

x

2

)

d

2

y

d

x

2

−

x

d

y

d

x

−

2

=

0

where  

y

=

(

sin

−

1

x

)

2

Using the result:

d

d

x

(

arcsin

x

)

=

1

√

1

−

x

2

In conjunction with the chain rule, then differentiating  

y

=

(

sin

−

1

x

)

2

wrt  

x

we have:

d

y

d

x

=

2

arcsin

x

√

1

−

x

2

And differentiating a second time, in conjunction with the quotient rule, we have:

d

2

y

d

x

2

=

(

√

1

−

x

2

)

(

2

√

1

−

x

2

)

−

(

1

2

(

−

2

x

)

√

1

−

x

2

)

(

2

arcsin

x

)

(

√

1

−

x

2

)

2

 

 

 

 

 

 

 

=

2

+

2

x

 

arcsin

x

√

1

−

x

2

1

−

x

2

And so, considering the LHS of the given expression:

L

H

S

=

(

1

−

x

2

)

d

2

y

d

x

2

−

x

d

y

d

x

−

2

 

 

 

 

 

 

 

 

=

(

1

−

x

2

)

⎧

⎪

⎨

⎪

⎩

2

+

2

x

 

arcsin

x

√

1

−

x

2

1

−

x

2

⎫

⎪

⎬

⎪

⎭

−

x

{

2

arcsin

x

√

1

−

x

2

}

−

2

 

 

 

 

 

 

 

 

=

2

+

2

x

 

arcsin

x

√

1

−

x

2

−

2

x

arcsin

x

√

1

−

x

2

−

2

 

 

 

 

 

 

 

 

=

0

 

 

QED

(

1

−

x

2

)

d

2

y

d

x

2

−

x

d

y

d

x

−

2

=

0

where  

y

=

(

sin

−

1

x

)

2

Using the result:

d

d

x

(

arcsin

x

)

=

1

√

1

−

x

2

In conjunction with the chain rule, then differentiating  

y

=

(

sin

−

1

x

)

2

wrt  

x

we have:

d

y

d

x

=

2

arcsin

x

√

1

−

x

2

And differentiating a second time, in conjunction with the quotient rule, we have:

d

2

y

d

x

2

=

(

√

1

−

x

2

)

(

2

√

1

−

x

2

)

−

(

1

2

(

−

2

x

)

√

1

−

x

2

)

(

2

arcsin

x

)

(

√

1

−

x

2

)

2

 

 

 

 

 

 

 

=

2

+

2

x

 

arcsin

x

√

1

−

x

2

1

−

x

2

And so, considering the LHS of the given expression:

L

H

S

=

(

1

−

x

2

)

d

2

y

d

x

2

−

x

d

y

d

x

−

2

 

 

 

 

 

 

 

 

=

(

1

−

x

2

)

⎧

⎪

⎨

⎪

⎩

2

+

2

x

 

arcsin

x

√

1

−

x

2

1

−

x

2

⎫

⎪

⎬

⎪

⎭

−

x

{

2

arcsin

x

√

1

−

x

2

}

−

2

 

 

 

 

 

 

 

 

=

2

+

2

x

 

arcsin

x

√

1

−

x

2

−

2

x

arcsin

x

√

1

−

x

2

−

2

 

 

 

 

 

 

 

 

=

0

 

 

QED

(

1

−

x

2

)

d

2

y

d

x

2

−

x

d

y

d

x

−

2

=

0

where  

y

=

(

sin

−

1

x

)

2

Using the result:

d

d

x

(

arcsin

x

)

=

1

√

1

−

x

2

In conjunction with the chain rule, then differentiating  

y

=

(

sin

−

1

x

)

2

wrt  

x

we have:

d

y

d

x

=

2

arcsin

x

√

1

−

x

2

And differentiating a second time, in conjunction with the quotient rule, we have:

d

2

y

d

x

2

=

(

√

1

−

x

2

)

(

2

√

1

−

x

2

)

−

(

1

2

(

−

2

x

)

√

1

−

x

2

)

(

2

arcsin

x

)

(

√

1

−

x

2

)

2

 

 

 

 

 

 

 

=

2

+

2

x

 

arcsin

x

√

1

−

x

2

1

−

x

2

And so, considering the LHS of the given expression:

L

H

S

=

(

1

−

x

2

)

d

2

y

d

x

2

−

x

d

y

d

x

−

2

 

 

 

 

 

 

 

 

=

(

1

−

x

2

)

⎧

⎪

⎨

⎪

⎩

2

+

2

x

 

arcsin

x

√

1

−

x

2

1

−

x

2

⎫

⎪

⎬

⎪

⎭

−

x

{

2

arcsin

x

√

1

−

x

2

}

−

2

 

 

 

 

 

 

 

 

=

2

+

2

x

 

arcsin

x

√

1

−

x

2

−

2

x

arcsin

x

√

1

−

x

2

−

2

 

 

 

 

 

 

 

 

=

0

 

 

QED

v

Step-by-step explanation: