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Popular >

y^'(t)=(3+y)(1-y),y(0)=2

Solution

y^{^{'}} (t) = (3 + y) (1 - y), y (0) = 2

Solution

3 - 2 y - y^{2}

Solution steps

Find

(3 + y) (1 - y)

given

y (0) = 2

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

y : 2

y : 2

= (3 + y) (1 - y)

Expand

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

= 3 - 2 y - y^{2}

Graph

Popular Examples

derivative of f(t)=2t^{-5}derivative of

f (t) = 2 t^{- 5}

y^{^{'}} + 2 y = \frac{1}{x}

y^{^{'}} + y = x^{2} y^{2}

derivative of log_{5}(x^2)

\frac{d}{d x} (lo g_{5} (x^{2}))

integral of (4w-11)/(2w^2+7w-4)

\int \frac{4 w - 11}{2 w ^{2} + 7 w - 4} d w

Frequently Asked Questions (FAQ)

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