What is the solution for y^{''}-5y^'-6y=-sin(2x) ?

The solution for y^{''}-5y^'-6y=-sin(2x) is y=c_{1}e^{6x}+c_{2}e^{-x}+1/20 sin(2x)-1/20 cos(2x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-5y^'-6y=-sin(2x)

y^{^{''}} - 5 y^{^{'}} - 6 y = - sin (2 x)

y = c_{1} e^{6 x} + c_{2} e^{- x} + \frac{1}{20} sin (2 x) - \frac{1}{20} cos (2 x)

Solution steps

y^{''} (x) - 5 y^{'} (x) - 6 y = - sin (2 x)

Solve linear ODE:

y = c_{1} e^{6 x} + c_{2} e^{- x} + \frac{1}{20} sin (2 x) - \frac{1}{20} cos (2 x)

y = c_{1} e^{6 x} + c_{2} e^{- x} + \frac{1}{20} sin (2 x) - \frac{1}{20} cos (2 x)

y^{^{''}} - 4 y^{^{'}} + 5 y = \frac{2 e ^{x}}{sin ( x )}

3 y^{^{''}} - 6 y^{^{'}} + 6 y = e x sec (x)

(D^{2} + 4) y = 5 e^{x} - 4 x^{2}

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

y^{^{''}} + 2 y = 3 x + 5

What is the solution for y^{''}-5y^'-6y=-sin(2x) ?
The solution for y^{''}-5y^'-6y=-sin(2x) is y=c_{1}e^{6x}+c_{2}e^{-x}+1/20 sin(2x)-1/20 cos(2x)