Question

How shall I proceed in this sum? If `x=sint` and `y=sin2t`, prove that (i) `(1-x^2)(dy/dx)^2=4(1-y^2)` (ii) `(1-x^2)(d^2(y)/dx^2)-xdy/dx+4y=0`

Answer 1

Given that `x=sint` and `y=sin2t`

Explanation:

Use `(dy)/dx = ((dy)/dt).((dt)/dx)`..........Chain Rule of differentiation

Here `(dy)/dt = 2cos2t` and `(dx)/dt = cost`

`rArr (dy)/dx = (2cos2t)/cost`

NOTE :
`rarr(dt)/dx = 1/[(dx)/dt ]`

`rarr 1-(sint)^2=(cost)^2`

`rarrsin2t =2sintcost`

`rarr cos2t=2(cost)^2-1`

For part (i) :-

Take LHS :-

`rArr(1-x^2)((dy)/dx)^2 = (1-(sint)^2).[(2cos2t)/cost]^2`

`rArr(1-x^2)((dy)/dx)^2 = cancel((cost)^2).(4(cos2t)^2)/cancel((cost)^2)=4(cos2t)^2`

Now take RHS :-

`4(1-y^2)=4(1-(sin2t)^2)=4(cos2t)^2`

Thus,

LHS `=` RHS `=4(cos2t)^2`

For part (ii) I am calculating only `(d^2(y))/dx^2` so you plug the values in the equation and similarly proceed as done in part (i)

{because yahaan type karna bahut lamba pad raha hai 😅😅😅}:-

`(d^2(y))/dx^2= (d[(dy)/dx])/dx`

`rArr(d^2(y))/dx^2= (d[(2cos2t)/cost])/dt`

`rArr(d^2(y))/dx^2= [cost(-4sin2t)-2cos2t.(-sint)]/(cost)^2`

`rArr(d^2(y))/dx^2= [-4cost (2sintcost)+2(2(cost)^2-1)sint] / (cost)^2`

`rArr(d^2(y))/dx^2= [-8sint.(cost)^2+4sint.(cost)^2-2sint]/ (cost)^2`

`rArr(d^2(y))/dx^2= [ -4sint.(cost)^2 -2sint]/ (cost)^2`

`:.(d^2(y))/dx^2= -4sint-2sect.tant`

Answer 2

`x = sin(t)`

`dx = cos(t)dt`

`y = sin(2t)`

`dy = 2cos(2t)dt`

`d^2y = -4sin(2t)dt^2`

(i)

`(1 - x^2)(dy/dx)^2 = (1 - sin^2(t))((2cos(2t)dt)/(cos(t)dt))^2`

`=cancel(cos^2(t))*(4cos^2(2t)cancel(dt))/cancel(cos^2(t))cancel(dt)`

`= 4(cos^2(2t))=4(1 - sin^2(2t)) = 4(1 - y^2)`

(ii)

`(1 - x^2)((d^2y)/dx^2) - x (dy/dx) + 4y`

`=(1 - sin^2(t))((-4sin(2t)dt^2)/(cos(t)dt)^2) - sin(t)(2cos(2t)dt)/(cos(t)dt) + 4sin(2t)`

`=cos^2(t)((-4sin(2t)dt^2)/(cos^2(t)dt^2)) - sin(t)(2cos(2t)dt)/(cos(t)dt) + 4sin(2t)`

`=cancel(cos^2(t))((-4sin(2t)cancel(dt^2))/(cancel(cos^2(t))cancel(dt^2))) - sin(t)(2cos(2t)cancel(dt))/(cos(t)cancel(dt)) + 4sin(2t)`

`=cancel(-4sin(2t)) + cancel(4sin(2t)) - ???`

Still in progress...