What is the solution for 1+y^2+2(t+1)y(dy)/(dt)=0,y(0)=1 ?

The solution for 1+y^2+2(t+1)y(dy)/(dt)=0,y(0)=1 is y=sqrt((-t+1)/(t+1))

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1+y^2+2(t+1)y(dy)/(dt)=0,y(0)=1

1 + y^{2} + 2 (t + 1) y \frac{d y}{d t} = 0, y (0) = 1

y = \frac{- t + 1}{t + 1}

Solution steps

1 + y^{2} + 2 (t + 1) y \frac{d y}{d t} = 0

Solve separable ODE:

y = \frac{- t + 1}{t + 1}

y = \frac{- t + 1}{t + 1}

x \frac{d y}{d x} + (1 - 3 x) y = e^{5 x}

f (x) = \frac{sin ( x )}{e ^{x}}

\int (x^{3} + 1.8 x^{2} - 2.4 x) d x

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

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What is the solution for 1+y^2+2(t+1)y(dy)/(dt)=0,y(0)=1 ?
The solution for 1+y^2+2(t+1)y(dy)/(dt)=0,y(0)=1 is y=sqrt((-t+1)/(t+1))