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Mathematics Equation Solving Wolfram Language

Find y' in terms of y?

Nathan Kilcrease

Posted 7 years ago

I've got the problem `x^2*Cos[y]^2-Sin[y]=0` and I need to find y' in terms of x and y, as well as y'' in terms of x, y, and y'. I understand how to find y' and y'' in terms of x, but trying to find it in terms of y and y' seems to be impossible. If you could explain the procedure to me, that would be great. Thanks!

POSTED BY: Nathan Kilcrease

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Carl Woll

Carl Woll, Wolfram Research

Posted 7 years ago

Does: Solve[Dt[x^2Cos[y]^2 - Sin[y] == 0, x], Dt[y, x]] ( {{Dt[y, x] -> (2 x Cos[y])/(1 + 2 x^2 Sin[y])}} ) Solve[Dt[x^2Cos[y]^2 - Sin[y] == 0, x, x], Dt[y, x, x]] (* {{Dt[y, {x, 2}] -> ( Cos[y] (2 - 2 x^2 Dt[y, x]^2 - 8 x Dt[y, x] Tan[y] + Dt[y, x]^2 Sec[y] Tan[y] + 2 x^2 Dt[y, x]^2 Tan[y]^2))/( 1 + 2 x^2 Sin[y])}} *) suffice?

Does:

Solve[Dt[x^2*Cos[y]^2 - Sin[y] == 0, x], Dt[y, x]]
(* {{Dt[y, x] -> (2 x Cos[y])/(1 + 2 x^2 Sin[y])}} *)

Solve[Dt[x^2*Cos[y]^2 - Sin[y] == 0, x, x], Dt[y, x, x]]
(* {{Dt[y, {x, 2}] -> (
   Cos[y] (2 - 2 x^2 Dt[y, x]^2 - 8 x Dt[y, x] Tan[y] + 
      Dt[y, x]^2 Sec[y] Tan[y] + 2 x^2 Dt[y, x]^2 Tan[y]^2))/(
   1 + 2 x^2 Sin[y])}} *)

suffice?

POSTED BY: Carl Woll

Nathan Kilcrease

Posted 7 years ago

I believe it does! Thank you so much!

POSTED BY: Nathan Kilcrease

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