What is the solution for y^{''}-y^'-6y=68cos(3x) ?

The solution for y^{''}-y^'-6y=68cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-34/39 sin(3x)-170/39 cos(3x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-y^'-6y=68cos(3x)

y^{^{''}} - y^{^{'}} - 6 y = 68 cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{34}{39} sin (3 x) - \frac{170}{39} cos (3 x)

Solution steps

y^{''} (x) - y^{'} (x) - 6 y = 68 cos (3 x)

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{34}{39} sin (3 x) - \frac{170}{39} cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{34}{39} sin (3 x) - \frac{170}{39} cos (3 x)

x^{^{''}} - 9 x - 1 = 0

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

y^{^{''}} + 9 y = cos (7 t)

(D^{2} + 3 D - 10) y = 20 + 5 e^{3 x}

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

What is the solution for y^{''}-y^'-6y=68cos(3x) ?
The solution for y^{''}-y^'-6y=68cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-34/39 sin(3x)-170/39 cos(3x)