What is the solution for y^{''}+4y^'+5=0 ?

The solution for y^{''}+4y^'+5=0 is y=c_{1}+c_{2}e^{-4t}-(5t)/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = c_{1} + c_{2} e^{- 4 t} - \frac{5 t}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 t} - \frac{5 t}{4}

y = c_{1} + c_{2} e^{- 4 t} - \frac{5 t}{4}

\int - 7 x^{6} d x

y = \frac{- 3 - 5 y}{- 8 y + 5 x}

(t^{2} - x t^{2}) \frac{d x}{d t} + x^{2} + t x^{2} = 0

\int_{0}^{1} x e^{- 3 x} d x

\int \frac{5 t}{1 - 25 t ^{4}} d t

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