en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y^'-5y=0

Solution

y^{^{''}} + 4 y^{^{'}} - 5 y = 0

Solution

y = c_{1} e^{t} + c_{2} e^{- 5 t}

Solution steps

y^{''} (t) + 4 y^{'} (t) - 5 y = 0

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{- 5 t}

y = c_{1} e^{t} + c_{2} e^{- 5 t}

Graph

Popular Examples

(x+2)^2(dy)/(dx)=5-8y-4xy

(x + 2)^{2} \frac{d y}{d x} = 5 - 8 y - 4 x y

limit as x approaches 3-of (-5)/(x-3)

x \to 3 - lim (\frac{- 5}{x - 3})

integral of (2+cos(x))^2

\int (2 + cos (x))^{2} d x

derivative of ((-x^2+2x+2))/((x+1)^4)derivative of

\frac{( - x ^{2} + 2 x + 2 )}{( x + 1 ) ^{4}}

y^{''}+y^'-6y=2e^{5x}

y^{^{''}} + y^{^{'}} - 6 y = 2 e^{5 x}

Frequently Asked Questions (FAQ)

What is the solution for y^{''}+4y^'-5y=0 ?
The solution for y^{''}+4y^'-5y=0 is y=c_{1}e^t+c_{2}e^{-5t}