What is the solution for y^'=((1-y)^2)/(x+1),y(0)=5 ?

The solution for y^'=((1-y)^2)/(x+1),y(0)=5 is y=-(4ln(x+1))/(-4ln(1+x)+1)+5/(-4ln(1+x)+1)

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y^'=((1-y)^2)/(x+1),y(0)=5

y^{^{'}} = \frac{( 1 - y ) ^{2}}{x + 1}, y (0) = 5

y = - \frac{4 ln ( x + 1 )}{- 4 ln ( 1 + x ) + 1} + \frac{5}{- 4 ln ( 1 + x ) + 1}

Solution steps

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

Solve separable ODE:

y = - \frac{4 ln ( x + 1 )}{- 4 ln ( 1 + x ) + 1} + \frac{5}{- 4 ln ( 1 + x ) + 1}

y = - \frac{4 ln ( x + 1 )}{- 4 ln ( 1 + x ) + 1} + \frac{5}{- 4 ln ( 1 + x ) + 1}

\frac{d}{d x} (ln (ln (sec (x))))

n = 1 \sum \infty 1 - \frac{7}{4 ^{n}}

\frac{\partial}{\partial x} (8 - x^{2} - y^{2})

24 y^{^{''}} - 2 y^{^{'}} - 15 y = 0

\frac{d}{d t} (cos^{2} (e^{c o s^{2} (t)}))

What is the solution for y^'=((1-y)^2)/(x+1),y(0)=5 ?
The solution for y^'=((1-y)^2)/(x+1),y(0)=5 is y=-(4ln(x+1))/(-4ln(1+x)+1)+5/(-4ln(1+x)+1)