en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

y = 5 e^{2 t}

Solution steps

y^{'} (t) = 2 y

Solve separable ODE:

y = 5 e^{2 t}

y = 5 e^{2 t}

Popular Examples

(dy)/(dx)=(y(5x-2))/(x(1-3y))

\frac{d y}{d x} = \frac{y ( 5 x - 2 )}{x ( 1 - 3 y )}

(dy)/(dx)=e^{3x-4y}

\frac{d y}{d x} = e^{3 x - 4 y}

y^'= t/(t^2y+y),y(0)=7

y^{^{'}} = \frac{t}{t ^{2} y + y}, y (0) = 7

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

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

(dy)/(dx)=(y+2)(x+1)

\frac{d y}{d x} = (y + 2) (x + 1)

Frequently Asked Questions (FAQ)

What is the solution for y^'=2y,y(0)=5 ?
The solution for y^'=2y,y(0)=5 is y=5e^{2t}