What is the y^'(t)=0.3y*(1-y/(2500))-y/4 ,y(0)=600 ?

The y^'(t)=0.3y*(1-y/(2500))-y/4 ,y(0)=600 is -0.072y+30

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y^'(t)=0.3y*(1-y/(2500))-y/4 ,y(0)=600

y^{^{'}} (t) = 0.3 y \cdot (1 - \frac{y}{2500}) - \frac{y}{4}, y (0) = 600

- 0.072 y + 30

Solution steps

Find

0.3 y \cdot (1 - \frac{y}{2500}) - \frac{y}{4}

given

y (0) = 600

0.3 y \cdot (1 - \frac{y}{2500}) - \frac{y}{4} = 0.3 y \cdot (1 - \frac{y}{2500}) - \frac{y}{4}

y : 600

\frac{y}{2500} : \frac{600}{2500}

\frac{y}{4} : \frac{600}{4}

= 0.3 \cdot 600 (1 - \frac{y}{2500}) - \frac{600}{4}

Expand

0.3 \cdot 600 (1 - \frac{y}{2500}) - \frac{600}{4} : - 0.072 y + 30

= - 0.072 y + 30

What is the y^'(t)=0.3y*(1-y/(2500))-y/4 ,y(0)=600 ?
The y^'(t)=0.3y*(1-y/(2500))-y/4 ,y(0)=600 is -0.072y+30