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

The (dy)/(dt)= 1/((y+2)^2),y(0)=1 is 1/9

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

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

\frac{1}{9}

Solution steps

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

Constant function results in a constant

y (0) = 1

= \frac{1}{( 1 + 2 ) ^{2}}

\frac{1}{( 1 + 2 ) ^{2}} = \frac{1}{9}

= \frac{1}{9}

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

\frac{d}{d x} (x^{5} e^{- 3 l n (x)})

\frac{d}{d x} ((x^{2} - 5 x)^{3})

\int \frac{2 x + 1}{( x + 1 ) ( x - 2 )} d x

\frac{d y}{d x} + 5 (tan (5 x)) y = 4 cos (5 x)

What is the (dy)/(dt)= 1/((y+2)^2),y(0)=1 ?
The (dy)/(dt)= 1/((y+2)^2),y(0)=1 is 1/9