What is the solution for y^{''}-y^'-2y=-2t+10t^2 ?

The solution for y^{''}-y^'-2y=-2t+10t^2 is y=c_{1}e^{2t}+c_{2}e^{-t}-5t^2+6t-8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-y^'-2y=-2t+10t^2

y^{^{''}} - y^{^{'}} - 2 y = - 2 t + 10 t^{2}

y = c_{1} e^{2 t} + c_{2} e^{- t} - 5 t^{2} + 6 t - 8

Solution steps

y^{''} (t) - y^{'} (t) - 2 y = - 2 t + 10 t^{2}

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- t} - 5 t^{2} + 6 t - 8

y = c_{1} e^{2 t} + c_{2} e^{- t} - 5 t^{2} + 6 t - 8

\int \frac{1}{5 x + 16} d x

y = 9 x^{2}, y = x^{2} + 7

h \to 0 lim (2 x h)

n = 5 \sum \infty \frac{1}{n ^{2}}

\int (\frac{1}{y} sin (\frac{x}{y}) + x) d x

What is the solution for y^{''}-y^'-2y=-2t+10t^2 ?
The solution for y^{''}-y^'-2y=-2t+10t^2 is y=c_{1}e^{2t}+c_{2}e^{-t}-5t^2+6t-8