Question

If `y=e^(msin^-1x)`, then show that `(1-x^2)y_2-xy_1=m^2y`?

Answer 1

To proceed we will need some standard Calculus results:

`d/dx e^(ax) = ae^(ax)`
`d/dx sin^(-1)x = 1/sqrt(1-x^2)`

Now we have:

`y = e^(msin^(-1)x)`

If we apply the chain rule then we get:

`y' = m \ e^(msin^(-1)x) * 1/sqrt(1-x^2)`
`\ \ \ \ = m \ e^(msin^(-1)x)/sqrt(1-x^2)`

And differentiating again and applying the quotient rule, along with the chain rule, we get:

`y'' = { (sqrt(1-x^2))(d/dxme^(msin^(-1)x)) - (me^(msin^(-1)x))(d/dxsqrt(1-x^2)) }/ (sqrt(1-x^2))^2`

`\ \ \ \ = { (sqrt(1-x^2))(m^2e^(msin^(-1)x)/sqrt(1-x^2)) - (me^(msin^(-1)x))(1/2(1-x^2)^(-1/2)*(-2x)) }/ (sqrt(1-x^2))^2`

`\ \ \ \ = { m^2e^(msin^(-1)x) + (mxe^(msin^(-1)x))/(sqrt(1-x^2)) }/ (1-x^2)`

`\ \ \ \ = { m^2y + xy' }/ (1-x^2)`

`:. (1-x^2)y'' = m^2y + xy'`

`:. (1-x^2)y'' -xy' = m^2y` QED

Answer 2

See the Explanation.

Explanation:

`y=e^(msin^-1x.`

Using the Chain Rule, we get,

`rArr y_1=dy/dx=(e^(msin^-1x))*d/dx(msin^-1x),`

`=y*{m/sqrt(1-x^2)},`

`rArr y_1*sqrt(1-x^2)=my.`

Squaring, `y_1^2(1-x^2)=m^2y^2`

Rediff,ing w.r.t. `x`, using the Product Rule and Chain Rule,

`y_1^2*d/dx(1-x^2)+(1-x^2)*d/dx(y_1^2)=m^2d/dx(y^2),`

`rArr y_1^2(-2x)+(1-x^2)(2y_1)*d/dx(y_1)=m^2(2y)*d/dx(y),`

`rArr y_1^2(-2x)+(1-x^2)(2y_1)(y_2)=m^2(2y)(y_1),`

Dividing by `2y_1!=0,` we get,

`(1-x^2)y_2-xy_1=m^2y.`

Hence, the Proof.

Enjoy Maths.!