What is the solution for y^{''}-2y^'-3y=cos(2x) ?

The solution for y^{''}-2y^'-3y=cos(2x) is y=c_{1}e^{3x}+c_{2}e^{-x}-4/65 sin(2x)-7/65 cos(2x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-2y^'-3y=cos(2x)

y^{^{''}} - 2 y^{^{'}} - 3 y = cos (2 x)

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{4}{65} sin (2 x) - \frac{7}{65} cos (2 x)

Solution steps

y^{''} (x) - 2 y^{'} (x) - 3 y = cos (2 x)

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{4}{65} sin (2 x) - \frac{7}{65} cos (2 x)

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{4}{65} sin (2 x) - \frac{7}{65} cos (2 x)

x^{^{''}} + 4 x = sin (5 t)

y^{^{''}} + y = sin (2 t), y (0) = 2, y^{^{'}} (0) = 1

y^{^{''}} + 9 y = tan (3 t)

y^{^{'''}} - 3 y^{^{''}} + 4 y = 4 e^{x} - 18 e^{- x}

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

What is the solution for y^{''}-2y^'-3y=cos(2x) ?
The solution for y^{''}-2y^'-3y=cos(2x) is y=c_{1}e^{3x}+c_{2}e^{-x}-4/65 sin(2x)-7/65 cos(2x)