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Popular Calculus >

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

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

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

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

integral from-1 to 1 of 11^x

\int_{- 1}^{1} 1 1^{x} d x

integral of x^{1/3}+6

\int x^{\frac{1}{3}} + 6 d x

derivative of-e^{-3x}

\frac{d}{d x} (- e^{- 3 x})

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