What is the solution for y^{''}+y^'+1.25=0 ?

The solution for y^{''}+y^'+1.25=0 is y=c_{1}+c_{2}e^{-t}-1.25t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+y^'+1.25=0

y^{^{''}} + y^{^{'}} + 1.25 = 0

y = c_{1} + c_{2} e^{- t} - 1.25 t

Solution steps

y^{''} (t) + y^{'} (t) + 1.25 = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- t} - 1.25 t

y = c_{1} + c_{2} e^{- t} - 1.25 t

(D^{2} - D - 2) y = 6 e - t^{2}

\frac{d ^{2} y}{d x ^{2}} - y = 19.2 e^{4 x} + 60 e^{x}

y^{^{''}} + 12 y^{^{'}} + 100 y = 48 sin (10 x)

y^{^{''}} - 16 y = 7 t, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + y^{^{'}} + 2 y - x - 3 = 0, y (0) = 0

What is the solution for y^{''}+y^'+1.25=0 ?
The solution for y^{''}+y^'+1.25=0 is y=c_{1}+c_{2}e^{-t}-1.25t