What is the solution for y^{''}+4y^'+29=0 ?

The solution for y^{''}+4y^'+29=0 is y=c_{1}+c_{2}e^{-4t}-(29t)/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y^'+29=0

y^{^{''}} + 4 y^{^{'}} + 29 = 0

y = c_{1} + c_{2} e^{- 4 t} - \frac{29 t}{4}

Solution steps

y^{''} (t) + 4 y^{'} (t) + 29 = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 t} - \frac{29 t}{4}

y = c_{1} + c_{2} e^{- 4 t} - \frac{29 t}{4}

y^{^{''}} - 5 y^{^{'}} + 6 y = 2 x + 3 e^{2 x}

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} - 3 y = 3 x e^{2 x}

y^{^{''}} - 2 y^{^{'}} + y = 3 x^{2} + 6 + 4 e x

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

y^{^{''}} + 9 y = 12 cos (3 x)

What is the solution for y^{''}+4y^'+29=0 ?
The solution for y^{''}+4y^'+29=0 is y=c_{1}+c_{2}e^{-4t}-(29t)/4