# Question 942da

Jan 2, 2017

See below.

#### Explanation:

Posing $y = {e}^{m \arccos \left(x\right)}$ we have

$y ' = - \frac{{e}^{m \arccos \left(x\right)} m}{\sqrt{1 - {x}^{2}}}$

y''=(e^(m arccos(x)) m^2)/( 1 - x^2)-(e^(m arccos(x)) mx)/(1 - x^2)^(3/2)

Substituting into the differential equation and simplifying, we can verify that it is identically satisfied.

A more simple approach is to consider $y = {e}^{m g \left(x\right)}$ and after substitution we get at

${e}^{m g \left(x\right)} m \left(m + x g ' \left(x\right) + m \left({x}^{2} - 1\right) g ' {\left(x\right)}^{2} + \left({x}^{2} - 1\right) g ' ' \left(x\right)\right) = 0$

now $g \left(x\right) = \arccos \left(x\right) \to g ' \left(x\right) = - \frac{1}{\sqrt{1 - {x}^{2}}} \to g ' ' \left(x\right) = - \frac{x}{1 - {x}^{2}} ^ \left(\frac{3}{2}\right)$

The substitution now is more direct, producing as expected

$\left(m + x g ' \left(x\right) + m \left({x}^{2} - 1\right) g ' {\left(x\right)}^{2} + \left({x}^{2} - 1\right) g ' ' \left(x\right)\right) = 0$

Feb 16, 2017

See the Proof In Explanation.

#### Explanation:

$y = {e}^{m {\cos}^{- 1} x}$

For the sake of brevity, we will use the notations ${y}_{1} \mathmr{and} {y}_{2}$ for

$\frac{\mathrm{dy}}{\mathrm{dx}} \mathmr{and} \frac{{d}^{2} y}{\mathrm{dx}} ^ 2$, resp.

Diff.ing the given eqn., using the Chain Rule, we get,

${y}_{1} = {e}^{m {\cos}^{-} 1 x} \left\{m {\cos}^{-} 1 x\right\} ' = y \left\{m \left(- \frac{1}{\sqrt{1 - {x}^{2}}}\right)\right\} .$

$\therefore {y}_{1} = \frac{- m y}{\sqrt{1 - {x}^{2}}} \Rightarrow \sqrt{1 - {x}^{2}} {y}_{1} = - m y .$

Squaring, $\left(1 - {x}^{2}\right) {y}_{1}^{2} = {m}^{2} {y}^{2.}$

Rediff.ing, using the Product Rule, we have,

$\left(1 - {x}^{2}\right) \left({y}_{1}^{2}\right) ' + {y}_{1}^{2} \left(1 - {x}^{2}\right) ' = {m}^{2} \left({y}^{2}\right) ' \ldots \ldots \ldots \ldots \ldots . \left(\star\right)$.

Now, note the following results derived by the Chain Rule , and

the Usual Rules of Diffn. These will be substd. in $\left(\star\right) .$

(1): (y_1^2)'=(2y_1)(y_1)'=2y_1y_2; (2): (y^2)'=(2y)(y)'=2yy_1;#

$\left(3\right) : \left(1 - {x}^{2}\right) ' = - 2 x .$

Therefore, by $\left(\star\right)$,

$2 \left(1 - {x}^{2}\right) {y}_{1} {y}_{2} - 2 x {y}_{1}^{2} = 2 {m}^{2} y {y}_{1.}$

Dividing throught by $2 {y}_{1} \ne 0$, we get the desired Proof :

$y = {e}^{m {\cos}^{-} 1 x} \Rightarrow \left(1 - {x}^{2}\right) \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 - 2 x \frac{\mathrm{dy}}{\mathrm{dx}} = {m}^{2} y .$

Enjoy Maths.!