en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y-y^'-2y=0

Solution

y - y^{^{'}} - 2 y = 0

Solution

y = e^{- t - c_{1}}

Solution steps

y - y^{'} (t) - 2 y = 0

Solve separable ODE:

y = e^{- t - c_{1}}

y = e^{- t - c_{1}}

Popular Examples

y^'=y^2xcos(x^2),y(pi/2)=1

y^{^{'}} = y^{2} x cos (x^{2}), y (\frac{π}{2}) = 1

y^'=2x(y^2+1),y(0)=0

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

(dx)/(dt)=e^{t+x}

\frac{d x}{d t} = e^{t + x}

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

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

xy^'+2y=3,xy(1)=5

x y^{^{'}} + 2 y = 3, x y (1) = 5

Frequently Asked Questions (FAQ)

What is the solution for y-y^'-2y=0 ?
The solution for y-y^'-2y=0 is y=e^{-t-c_{1}}