What is the solution for y^{''}-4y^'-5y=50t^3 ?

The solution for y^{''}-4y^'-5y=50t^3 is y=c_{1}e^{5t}+c_{2}e^{-t}-10t^3+24t^2-(252t)/5+1248/25

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-4y^'-5y=50t^3

y^{^{''}} - 4 y^{^{'}} - 5 y = 50 t^{3}

y = c_{1} e^{5 t} + c_{2} e^{- t} - 10 t^{3} + 24 t^{2} - \frac{252 t}{5} + \frac{1248}{25}

Solution steps

y^{''} (t) - 4 y^{'} (t) - 5 y = 50 t^{3}

Solve linear ODE:

y = c_{1} e^{5 t} + c_{2} e^{- t} - 10 t^{3} + 24 t^{2} - \frac{252 t}{5} + \frac{1248}{25}

y = c_{1} e^{5 t} + c_{2} e^{- t} - 10 t^{3} + 24 t^{2} - \frac{252 t}{5} + \frac{1248}{25}

\int \frac{e ^{- x} - e ^{2 x}}{2} d x

\frac{d}{d x} (\frac{x - 3}{x + 3})

\frac{d}{d x} (\frac{m}{x})

f (t) = \frac{2 t}{t ^{2} + 1}

f (x) = \frac{1}{x ^{2}}, at x = 5

What is the solution for y^{''}-4y^'-5y=50t^3 ?
The solution for y^{''}-4y^'-5y=50t^3 is y=c_{1}e^{5t}+c_{2}e^{-t}-10t^3+24t^2-(252t)/5+1248/25