What is the solution for θ^{''}(t)-θ(t)=tsin(t) ?

The solution for θ^{''}(t)-θ(t)=tsin(t) is θ(t)=c_{1}e^t+c_{2}e^{-t}-(tsin(t))/2-1/2 cos(t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

θ^{''}(t)-θ(t)=tsin(t)

θ^{^{''}} - θ = t sin (t)

θ (t) = c_{1} e^{t} + c_{2} e^{- t} - \frac{t sin ( t )}{2} - \frac{1}{2} cos (t)

Solution steps

θ^{''} (t) - θ (t) = t sin (t)

Solve linear ODE:

θ (t) = c_{1} e^{t} + c_{2} e^{- t} - \frac{t sin ( t )}{2} - \frac{1}{2} cos (t)

θ (t) = c_{1} e^{t} + c_{2} e^{- t} - \frac{t sin ( t )}{2} - \frac{1}{2} cos (t)

What is the solution for θ^{''}(t)-θ(t)=tsin(t) ?
The solution for θ^{''}(t)-θ(t)=tsin(t) is θ(t)=c_{1}e^t+c_{2}e^{-t}-(tsin(t))/2-1/2 cos(t)