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Popular >

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}

Popular Examples

2yt(dy)+y^2(dt)=0

2 y t (d y) + y^{2} (d t) = 0

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

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

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

y^{^{'}} = \frac{y ^{2}}{x ^{3}}, y (2) = 4

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

x y y^{^{'}} = (x + 1) (y + 1)

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

(3 y^{2} - 4 x^{- 1} y) d x + (3 x y - 4) d y = 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