en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular Calculus >

y^'(t)=(3+y)(1-y),y(0)=4

Solution

y^{^{'}} (t) = (3 + y) (1 - y), y (0) = 4

Solution

3 - 2 y - y^{2}

Solution steps

Find

(3 + y) (1 - y)

given

y (0) = 4

(3 + y) (1 - y) = (3 + y) (1 - y)

y : 4

y : 4

= (3 + y) (1 - y)

Expand

(3 + y) (1 - y) : 3 - 2 y - y^{2}

= 3 - 2 y - y^{2}

Graph

Popular Examples

y^'+y=y(te^{t^2}+1),y(0)=1

y^'=(y^2)/(x^3),y(2)=4

xyy^'=(x+1)(y+1)

(3y^2-4x^{-1}y)dx+(3xy-4)dy=0

Frequently Asked Questions (FAQ)

What is the y^'(t)=(3+y)(1-y),y(0)=4 ?
The y^'(t)=(3+y)(1-y),y(0)=4 is 3-2y-y^2