What is the solution for y^{''}-y^'-6y=45cos(3x) ?

The solution for y^{''}-y^'-6y=45cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-15/26 sin(3x)-75/26 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=45cos(3x)

y^{^{''}} - y^{^{'}} - 6 y = 45 cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{15}{26} sin (3 x) - \frac{75}{26} cos (3 x)

Solution steps

y^{''} (x) - y^{'} (x) - 6 y = 45 cos (3 x)

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{15}{26} sin (3 x) - \frac{75}{26} cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{15}{26} sin (3 x) - \frac{75}{26} cos (3 x)

y^{^{''}} + 4 y^{^{'}} + 3 y = 8 e^{- 2 x}

y^{^{''}} + 3 y^{^{'}} + 2 y = 3 e^{2 t}

y^{^{''}} - 8 y^{^{'}} + 16 y = e^{- x}

y^{^{'''}} - 3 y^{^{''}} + 2 y^{^{'}} = 12 e^{2 x} + 24 x^{2}

\frac{d ^{2} y}{d t ^{2}} - y = e^{t}, y (0) = 1, y^{^{'}} (0) = 1

What is the solution for y^{''}-y^'-6y=45cos(3x) ?
The solution for y^{''}-y^'-6y=45cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-15/26 sin(3x)-75/26 cos(3x)