What is the y^'(t)= 1/8 y(1-y/8),y(0)=60 ?

The y^'(t)= 1/8 y(1-y/8),y(0)=60 is 15/2-(15y)/(16)

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y^'(t)= 1/8 y(1-y/8),y(0)=60

y^{^{'}} (t) = \frac{1}{8} y (1 - \frac{y}{8}), y (0) = 60

\frac{15}{2} - \frac{15 y}{16}

Solution steps

Find

\frac{1}{8} y (1 - \frac{y}{8})

given

y (0) = 60

\frac{1}{8} y (1 - \frac{y}{8}) = \frac{1}{8} y (1 - \frac{y}{8})

y : 60

\frac{y}{8} : \frac{60}{8}

= \frac{1}{8} \cdot 60 (1 - \frac{y}{8})

Expand

\frac{1}{8} \cdot 60 (1 - \frac{y}{8}) : \frac{15}{2} - \frac{15 y}{16}

= \frac{15}{2} - \frac{15 y}{16}

\frac{d y}{d x} = \frac{- 0.3}{cos ( y )}, y (0) = \frac{π}{4}

y^{^{'}} = \frac{x ^{3}}{3 e ^{y + 1}}

(ln (y))^{3} (\frac{d y}{d x}) = x^{3} y, y (1) = e^{2}

\frac{d y}{d x} = k 2 y + a

(y^{^{'}}) = e^{x} e^{y}

What is the y^'(t)= 1/8 y(1-y/8),y(0)=60 ?
The y^'(t)= 1/8 y(1-y/8),y(0)=60 is 15/2-(15y)/(16)