What is the solution for y^{''}+4y^'=2sin(2x) ?

The solution for y^{''}+4y^'=2sin(2x) is y=c_{1}+c_{2}e^{-4x}-1/10 sin(2x)-1/5 cos(2x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = c_{1} + c_{2} e^{- 4 x} - \frac{1}{10} sin (2 x) - \frac{1}{5} cos (2 x)

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 x} - \frac{1}{10} sin (2 x) - \frac{1}{5} cos (2 x)

y = c_{1} + c_{2} e^{- 4 x} - \frac{1}{10} sin (2 x) - \frac{1}{5} cos (2 x)

y^{^{''}} - y^{^{'}} - 6 y = 6 t^{2}

y^{^{''}} - 3 y^{^{'}} = 4 sin (x)

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

y^{^{''}} - 2 y^{^{'}} - 3 y = 6 x e^{3 x}

y^{^{''}} + 4 y^{^{'}} + 3 y = e^{5 x}

What is the solution for y^{''}+4y^'=2sin(2x) ?
The solution for y^{''}+4y^'=2sin(2x) is y=c_{1}+c_{2}e^{-4x}-1/10 sin(2x)-1/5 cos(2x)