What is the solution for y^{''}+4y^'+y=6sin(x) ?

The solution for y^{''}+4y^'+y=6sin(x) is y=c_{1}e^{(-2+sqrt(3))x}+c_{2}e^{(-2-sqrt(3))x}-3/2 cos(x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y^'+y=6sin(x)

y^{^{''}} + 4 y^{^{'}} + y = 6 sin (x)

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

Solution steps

y^{''} (x) + 4 y^{'} (x) + y = 6 sin (x)

Solve linear ODE:

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

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

x y^{^{''}} + x y^{^{'}} = x

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

y^{^{''}} + 16 y = 2 cos (4 t)

y^{^{'''}} - 2 y^{^{''}} - y^{^{'}} + 2 y = e^{5 x}

y^{^{''}} + 6 y^{^{'}} + 9 y = x

What is the solution for y^{''}+4y^'+y=6sin(x) ?
The solution for y^{''}+4y^'+y=6sin(x) is y=c_{1}e^{(-2+sqrt(3))x}+c_{2}e^{(-2-sqrt(3))x}-3/2 cos(x)