en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=k(y+2),y(0)=5

Solution

y^{^{'}} = k (y + 2), y (0) = 5

Solution

y = 7 e^{k t} - 2

Solution steps

y^{'} (t) = k (y + 2)

Solve separable ODE:

y = 7 e^{k t} - 2

y = 7 e^{k t} - 2

Popular Examples

(dy)/(y^{1/2)}=xdx

\frac{d y}{y ^{\frac{1}{2}}} = x d x

(dy)/(dx)=(y^2)/(sqrt(x))

\frac{d y}{d x} = \frac{y ^{2}}{x}

x^'=1-3x+2x^2,x(0)=2

x^{^{'}} = 1 - 3 x + 2 x^{2}, x (0) = 2

3y+5+(3x-7)(dy)/(dx)=0

3 y + 5 + (3 x - 7) \frac{d y}{d x} = 0

y^'=(7x^2-1)/(7+4y)

y^{^{'}} = \frac{7 x ^{2} - 1}{7 + 4 y}

Frequently Asked Questions (FAQ)

What is the solution for y^'=k(y+2),y(0)=5 ?
The solution for y^'=k(y+2),y(0)=5 is y=7e^{kt}-2