What is the solution for x^{''}+4x=12cos(t) ?

The solution for x^{''}+4x=12cos(t) is x=c_{1}cos(2t)+c_{2}sin(2t)+4cos(t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{''}+4x=12cos(t)

x^{^{''}} + 4 x = 12 cos (t)

x = c_{1} cos (2 t) + c_{2} sin (2 t) + 4 cos (t)

Solution steps

x^{''} (t) + 4 x = 12 cos (t)

Solve linear ODE:

x = c_{1} cos (2 t) + c_{2} sin (2 t) + 4 cos (t)

x = c_{1} cos (2 t) + c_{2} sin (2 t) + 4 cos (t)

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 2 y = x^{2}

y^{^{''}} + 4 y = 4598 cos (2 x)

x^{^{''}} - 2 x^{^{'}} + 2 x = e^{- t} cos (2 t)

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

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} + y = x

What is the solution for x^{''}+4x=12cos(t) ?
The solution for x^{''}+4x=12cos(t) is x=c_{1}cos(2t)+c_{2}sin(2t)+4cos(t)