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#
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Given that #x=sint# and #y=sin2t#
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#
#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...
