What is the solution for y^'=(1+2y)/(sqrt(4-x^2)),y(1)=0 ?

The solution for y^'=(1+2y)/(sqrt(4-x^2)),y(1)=0 is y=(e^{(6arcsin(x/2)-pi)/3})/(2)-1/2

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y^'=(1+2y)/(sqrt(4-x^2)),y(1)=0

y^{^{'}} = \frac{1 + 2 y}{4 - x ^{2}}, y (1) = 0

y = \frac{e ^{\frac{6 a r c s i n ( \frac{x}{2} ) - π}{3}}}{2} - \frac{1}{2}

Solution steps

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

Solve separable ODE:

y = \frac{e ^{\frac{6 a r c s i n ( \frac{x}{2} ) - π}{3}}}{2} - \frac{1}{2}

y = \frac{e ^{\frac{6 a r c s i n ( \frac{x}{2} ) - π}{3}}}{2} - \frac{1}{2}

\frac{d v}{d t} = 5 - 4 v

\frac{d y}{d x} y = sinh (11 x)

\frac{1}{y + y} d y = \frac{1}{x + x} d x

d y + sin (x) y d x - sin (x) d x = 0

\frac{( d y )}{( d t )} = 2 t cos^{(2)} (y)

What is the solution for y^'=(1+2y)/(sqrt(4-x^2)),y(1)=0 ?
The solution for y^'=(1+2y)/(sqrt(4-x^2)),y(1)=0 is y=(e^{(6arcsin(x/2)-pi)/3})/(2)-1/2