Question

Question #942da

Answer 1

See below.

Explanation:

Posing `y=e^(m arccos(x))` we have

`y'=-(e^(m arccos(x)) 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(x))` and after substitution we get at

`e^(m g(x))m(m + x g'(x)+m(x^2-1)g'(x)^2+(x^2-1)g''(x))=0`

now `g(x) = arccos(x)->g'(x)=-1/sqrt[1 - x^2]->g''(x)=-x/(1 - x^2)^(3/2)`

The substitution now is more direct, producing as expected

`(m + x g'(x)+m(x^2-1)g'(x)^2+(x^2-1)g''(x))=0`

Answer 2

See the Proof In Explanation.

Explanation:

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

For the sake of brevity, we will use the notations `y_1 and y_2` for

`dy/dx and (d^2y)/dx^2`, resp.

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

`y_1=e^(mcos^-1x){mcos^-1x}'=y{m(-1/sqrt(1-x^2))}.`

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

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

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

`(1-x^2)(y_1^2)'+y_1^2(1-x^2)'=m^2(y^2)'................(star)`.

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

the Usual Rules of Diffn. These will be substd. in `(star).`

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

`(3): (1-x^2)'=-2x.`

Therefore, by `(star)`,

`2(1-x^2)y_1y_2-2xy_1^2=2m^2yy_1.`

Dividing throught by `2y_1ne0`, we get the desired Proof :

`y=e^(mcos^-1x) rArr (1-x^2)(d^2y)/dx^2-2xdy/dx=m^2y.`

Enjoy Maths.!