Let x = cost, transform the equation d ^ 2Y / DX ^ 2-x / (1-x ^ 2) * dy / DX + Y / (1-x ^ 2) = 0 The answer is d ^ 2Y / dt ^ 2 + y = 0. I want to see the solution

Let x = cost, transform the equation d ^ 2Y / DX ^ 2-x / (1-x ^ 2) * dy / DX + Y / (1-x ^ 2) = 0 The answer is d ^ 2Y / dt ^ 2 + y = 0. I want to see the solution

D ^ 2Y / DX ^ 2 = D (dy / DX) / DX = D (- dy / (sintdt)) / (- sintdt) = ((d ^ 2Y / dt * Sint dy / dt * cost) / (Sint) ^ 2) DT / (- sintdt) = D ^ 2Y / dt ^ 2 / (Sint) ^ 2-dy / dt * cost / (Sint) ^ 3 the original equation can be reduced to 1 / (Sint) ^ 2 * d ^ 2Y / dt ^ 2-cost / (Sint ^ 3) * dy / dt + cost / (Sint)

Given x = t (1-cost) and y = tcost, y = f (x) is determined to find dy / DX and d ^ 2Y / DX ^ 2,

Given x = t (1-cost) and y = tcost, y = f (x) is determined to find dy / DX and D ² y/dx ² y'=dy/dx=(dy/dt)/(dx/dt)=(cost-tsint)/(1-cost+tsint)； y''=d ² y/dx ²= dy'/dx=(dy'/dt)/(dx/dt)={[(1-cost+tsint)(-sint-sint-tcost)-(c...

How to find the second derivative of the derivative dy / DX of the parametric equation x = a (t-sint) y = a (1-cost)?

obviously
dx/dt=a(1-cost)
dy/dt=a*sint
that
dy/dx=sint /(1-cost)
Continue to find the second derivative
d(dy/dx)/dt *dt/dx
=[(sint)' *(1-cost) -sint *(1-cost)']/(1-cost)^2 *1/ a(1-cost)
=(cost-1)/(1-cost)^2 *1/ a(1-cost)
= -1/ [a(1-cost)^2]

Y = e ^ t * cost, x = e ^ t * Sint, find y ` `. Why is the result I get with y ` ` = (y ` ` different from that of Y ` ` = D (dy / DX) / DX?

Sorry to blow! These two methods are wrong! (y ')' is wrong. For the parametric equation, use the second method

X = a (cost + TSINT) y = a (sint tcost) derivation dy / DX Which great God taught me

analysis
x=acost+atsint
y=asint-atcost
dx=-asint+asint+atcost
dy=acost-acost+atsint
∴dy/dx=(acost-acost+asint)/(atcost)
=asint/atcost
=tan/t
Typing decisively,

If {x = 7 (t-sint), y = 7 (1-cost), then dy / DX=

dx=(7-7cost)dt dy=(7sint)dt dy/dx=(7sint)/(7-7cost)