What is the solution for x^{''}+5x^'+4x=t^2+3t+5 ?

The solution for x^{''}+5x^'+4x=t^2+3t+5 is x=c_{1}e^{-t}+c_{2}e^{-4t}+(t^2)/4+t/8+31/32

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{''}+5x^'+4x=t^2+3t+5

x^{^{''}} + 5 x^{^{'}} + 4 x = t^{2} + 3 t + 5

x = c_{1} e^{- t} + c_{2} e^{- 4 t} + \frac{t ^{2}}{4} + \frac{t}{8} + \frac{31}{32}

Solution steps

x^{''} (t) + 5 x^{'} (t) + 4 x = t^{2} + 3 t + 5

Solve linear ODE:

x = c_{1} e^{- t} + c_{2} e^{- 4 t} + \frac{t ^{2}}{4} + \frac{t}{8} + \frac{31}{32}

x = c_{1} e^{- t} + c_{2} e^{- 4 t} + \frac{t ^{2}}{4} + \frac{t}{8} + \frac{31}{32}

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

y^{^{''}} + 16 y = e^{3 t}

y^{^{''}} + 16 y = 2 cos (4 x)

y^{^{''}} - 10 y^{^{'}} + 25 y = 48 e^{x}

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

What is the solution for x^{''}+5x^'+4x=t^2+3t+5 ?
The solution for x^{''}+5x^'+4x=t^2+3t+5 is x=c_{1}e^{-t}+c_{2}e^{-4t}+(t^2)/4+t/8+31/32