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Popular >

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

Solution

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

Solution

y^{2} - 1

Solution steps

Find

(y + 1) (y - 1)

given

y (0) = 3

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

y : 3

y : 3

= (y + 1) (y - 1)

Expand

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

= y^{2} - 1

Graph

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\frac{\partial}{\partial u} (u^{2} + v^{2} - 4)

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\int x \cdot (x^{2} - 3) d x

Frequently Asked Questions (FAQ)

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