What is the solution for y^'= 1/((t*y+y+t+1)),y(0)=-1 ?

The solution for y^'= 1/((t*y+y+t+1)),y(0)=-1 is y=-1+sqrt(2)sqrt(ln(t+1)),y=-sqrt(2)sqrt(ln(t+1))-1

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y^'= 1/((t*y+y+t+1)),y(0)=-1

y^{^{'}} = \frac{1}{( t \cdot y + y + t + 1 )}, y (0) = - 1

y = - 1 + 2 ln (t + 1), y = - 2 ln (t + 1) - 1

Solution steps

y^{'} (t) = \frac{1}{( t \cdot y + y + t + 1 )}

Solve separable ODE:

y = - 1 + 2 ln (t + 1), y = - 2 ln (t + 1) - 1

y = - 1 + 2 ln (t + 1), y = - 2 ln (t + 1) - 1

f (x) = x + 35, (1, 6)

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

e^{- t} cos (4 t)

\int (3 x^{2} + 2 x) e^{(x^{3} + x^{2})} d x

f (x) = - x^{3} - 4 x - 3, at x = - 1

What is the solution for y^'= 1/((t*y+y+t+1)),y(0)=-1 ?
The solution for y^'= 1/((t*y+y+t+1)),y(0)=-1 is y=-1+sqrt(2)sqrt(ln(t+1)),y=-sqrt(2)sqrt(ln(t+1))-1