What is the solution for y(4-x^2)^{1/2}dy=(4+y^2)^{1/2}dx ?

The solution for y(4-x^2)^{1/2}dy=(4+y^2)^{1/2}dx is

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y(4-x^2)^{1/2}dy=(4+y^2)^{1/2}dx

y (4 - x^{2})^{\frac{1}{2}} d y = (4 + y^{2})^{\frac{1}{2}} d x

y = \frac{- 8 + c _{1}^{2} + arcsin ^{2} ( \frac{x}{2} ) + 2 c _{1} arcsin ( \frac{x}{2} ) + ( arcsin ^{4} ( \frac{x}{2} ) + 4 c _{1} arcsin ^{3} ( \frac{x}{2} ) + 6 c _{1}^{2} arcsin ^{2} ( \frac{x}{2} ) + 4 c _{1}^{3} arcsin ( \frac{x}{2} ) + c _{1}^{4} ) ^{\frac{1}{2}}}{2}, y = - \frac{- 8 + c _{1}^{2} + arcsin ^{2} ( \frac{x}{2} ) + 2 c _{1} arcsin ( \frac{x}{2} ) + ( arcsin ^{4} ( \frac{x}{2} ) + 4 c _{1} arcsin ^{3} ( \frac{x}{2} ) + 6 c _{1}^{2} arcsin ^{2} ( \frac{x}{2} ) + 4 c _{1}^{3} arcsin ( \frac{x}{2} ) + c _{1}^{4} ) ^{\frac{1}{2}}}{2}, y = \frac{- 8 + c _{1}^{2} + arcsin ^{2} ( \frac{x}{2} ) + 2 c _{1} arcsin ( \frac{x}{2} ) - ( arcsin ^{4} ( \frac{x}{2} ) + 4 c _{1} arcsin ^{3} ( \frac{x}{2} ) + 6 c _{1}^{2} arcsin ^{2} ( \frac{x}{2} ) + 4 c _{1}^{3} arcsin ( \frac{x}{2} ) + c _{1}^{4} ) ^{\frac{1}{2}}}{2}, y = - \frac{- 8 + c _{1}^{2} + arcsin ^{2} ( \frac{x}{2} ) + 2 c _{1} arcsin ( \frac{x}{2} ) - ( arcsin ^{4} ( \frac{x}{2} ) + 4 c _{1} arcsin ^{3} ( \frac{x}{2} ) + 6 c _{1}^{2} arcsin ^{2} ( \frac{x}{2} ) + 4 c _{1}^{3} arcsin ( \frac{x}{2} ) + c _{1}^{4} ) ^{\frac{1}{2}}}{2}

Solution steps

y (4 - x^{2})^{\frac{1}{2}} d y = (4 + y^{2})^{\frac{1}{2}} d x

Solve separable ODE:

y = \frac{- 8 + c _{1}^{2} + arcsin ^{2} ( \frac{x}{2} ) + 2 c _{1} arcsin ( \frac{x}{2} ) + ( arcsin ^{4} ( \frac{x}{2} ) + 4 c _{1} arcsin ^{3} ( \frac{x}{2} ) + 6 c _{1}^{2} arcsin ^{2} ( \frac{x}{2} ) + 4 c _{1}^{3} arcsin ( \frac{x}{2} ) + c _{1}^{4} ) ^{\frac{1}{2}}}{2}, y = - \frac{- 8 + c _{1}^{2} + arcsin ^{2} ( \frac{x}{2} ) + 2 c _{1} arcsin ( \frac{x}{2} ) + ( arcsin ^{4} ( \frac{x}{2} ) + 4 c _{1} arcsin ^{3} ( \frac{x}{2} ) + 6 c _{1}^{2} arcsin ^{2} ( \frac{x}{2} ) + 4 c _{1}^{3} arcsin ( \frac{x}{2} ) + c _{1}^{4} ) ^{\frac{1}{2}}}{2}, y = \frac{- 8 + c _{1}^{2} + arcsin ^{2} ( \frac{x}{2} ) + 2 c _{1} arcsin ( \frac{x}{2} ) - ( arcsin ^{4} ( \frac{x}{2} ) + 4 c _{1} arcsin ^{3} ( \frac{x}{2} ) + 6 c _{1}^{2} arcsin ^{2} ( \frac{x}{2} ) + 4 c _{1}^{3} arcsin ( \frac{x}{2} ) + c _{1}^{4} ) ^{\frac{1}{2}}}{2}, y = - \frac{- 8 + c _{1}^{2} + arcsin ^{2} ( \frac{x}{2} ) + 2 c _{1} arcsin ( \frac{x}{2} ) - ( arcsin ^{4} ( \frac{x}{2} ) + 4 c _{1} arcsin ^{3} ( \frac{x}{2} ) + 6 c _{1}^{2} arcsin ^{2} ( \frac{x}{2} ) + 4 c _{1}^{3} arcsin ( \frac{x}{2} ) + c _{1}^{4} ) ^{\frac{1}{2}}}{2}

y = \frac{- 8 + c _{1}^{2} + arcsin ^{2} ( \frac{x}{2} ) + 2 c _{1} arcsin ( \frac{x}{2} ) + ( arcsin ^{4} ( \frac{x}{2} ) + 4 c _{1} arcsin ^{3} ( \frac{x}{2} ) + 6 c _{1}^{2} arcsin ^{2} ( \frac{x}{2} ) + 4 c _{1}^{3} arcsin ( \frac{x}{2} ) + c _{1}^{4} ) ^{\frac{1}{2}}}{2}, y = - \frac{- 8 + c _{1}^{2} + arcsin ^{2} ( \frac{x}{2} ) + 2 c _{1} arcsin ( \frac{x}{2} ) + ( arcsin ^{4} ( \frac{x}{2} ) + 4 c _{1} arcsin ^{3} ( \frac{x}{2} ) + 6 c _{1}^{2} arcsin ^{2} ( \frac{x}{2} ) + 4 c _{1}^{3} arcsin ( \frac{x}{2} ) + c _{1}^{4} ) ^{\frac{1}{2}}}{2}, y = \frac{- 8 + c _{1}^{2} + arcsin ^{2} ( \frac{x}{2} ) + 2 c _{1} arcsin ( \frac{x}{2} ) - ( arcsin ^{4} ( \frac{x}{2} ) + 4 c _{1} arcsin ^{3} ( \frac{x}{2} ) + 6 c _{1}^{2} arcsin ^{2} ( \frac{x}{2} ) + 4 c _{1}^{3} arcsin ( \frac{x}{2} ) + c _{1}^{4} ) ^{\frac{1}{2}}}{2}, y = - \frac{- 8 + c _{1}^{2} + arcsin ^{2} ( \frac{x}{2} ) + 2 c _{1} arcsin ( \frac{x}{2} ) - ( arcsin ^{4} ( \frac{x}{2} ) + 4 c _{1} arcsin ^{3} ( \frac{x}{2} ) + 6 c _{1}^{2} arcsin ^{2} ( \frac{x}{2} ) + 4 c _{1}^{3} arcsin ( \frac{x}{2} ) + c _{1}^{4} ) ^{\frac{1}{2}}}{2}

x y^{^{'}} = y tan (ln (y))

\frac{d y}{d t} = - \frac{t}{2 y}

(3 x y^{2} - x^{3}) \frac{d y}{d x} = 3 y^{3} - x^{2} y

\frac{d y}{d x} = \frac{sin ( 3 x )}{cos ( 4 y )}

y^{^{'}} = \frac{x ^{3}}{y}

What is the solution for y(4-x^2)^{1/2}dy=(4+y^2)^{1/2}dx ?
The solution for y(4-x^2)^{1/2}dy=(4+y^2)^{1/2}dx is