What is the solution for x^{''}-3x^'+2x=20sin(2t) ?

The solution for x^{''}-3x^'+2x=20sin(2t) is x=c_{1}e^{2t}+c_{2}e^t-sin(2t)+3cos(2t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{''}-3x^'+2x=20sin(2t)

x^{^{''}} - 3 x^{^{'}} + 2 x = 20 sin (2 t)

x = c_{1} e^{2 t} + c_{2} e^{t} - sin (2 t) + 3 cos (2 t)

Solution steps

x^{''} (t) - 3 x^{'} (t) + 2 x = 20 sin (2 t)

Solve linear ODE:

x = c_{1} e^{2 t} + c_{2} e^{t} - sin (2 t) + 3 cos (2 t)

x = c_{1} e^{2 t} + c_{2} e^{t} - sin (2 t) + 3 cos (2 t)

y^{^{''}} - 12 y^{^{'}} + 36 y = 8 e^{6 x}

x^{^{''}} - 5 x^{^{'}} + 6 x = 12, x (0) = 2, x^{^{'}} (0) = 0

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

y^{^{''}} - 2 y^{^{'}} + y = e^{t - 1}, y (1) = 0, y^{^{'}} (1) = 5

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

What is the solution for x^{''}-3x^'+2x=20sin(2t) ?
The solution for x^{''}-3x^'+2x=20sin(2t) is x=c_{1}e^{2t}+c_{2}e^t-sin(2t)+3cos(2t)