en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

y^{2} - 2 y - 3

Solution steps

Find

(y + 1) (y - 3)

given

y (0) = 1

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

y : 1

y : 1

= (y + 1) (y - 3)

Expand

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

= y^{2} - 2 y - 3

Popular Examples

t^3(dy)/(dt)+3t^2y=7cos(t),y(pi)=0

t^{3} \frac{d y}{d t} + 3 t^{2} y = 7 cos (t), y (π) = 0

limit as x approaches infinity of-x/s

x \to \infty lim (- \frac{x}{s})

derivative of 18(2e^{x^2}x^2+e^{x^2})derivative of

18 (2 e^{x^{2}} x^{2} + e^{x^{2}})

longdivision (x-3)/(x+2)long division

\frac{x - 3}{x + 2}

derivative of 3^{2x^2}

\frac{d}{d x} (3^{2 x^{2}})

Frequently Asked Questions (FAQ)

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