What is the solution for y^{''}-7y^'+10y=-2cos(2t) ?

The solution for y^{''}-7y^'+10y=-2cos(2t) is y=c_{1}e^{5t}+c_{2}e^{2t}+7/58 sin(2t)-3/58 cos(2t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}-7y^'+10y=-2cos(2t)

y^{^{''}} - 7 y^{^{'}} + 10 y = - 2 cos (2 t)

y = c_{1} e^{5 t} + c_{2} e^{2 t} + \frac{7}{58} sin (2 t) - \frac{3}{58} cos (2 t)

Solution steps

y^{''} (t) - 7 y^{'} (t) + 10 y = - 2 cos (2 t)

Solve linear ODE:

y = c_{1} e^{5 t} + c_{2} e^{2 t} + \frac{7}{58} sin (2 t) - \frac{3}{58} cos (2 t)

y = c_{1} e^{5 t} + c_{2} e^{2 t} + \frac{7}{58} sin (2 t) - \frac{3}{58} cos (2 t)

What is the solution for y^{''}-7y^'+10y=-2cos(2t) ?
The solution for y^{''}-7y^'+10y=-2cos(2t) is y=c_{1}e^{5t}+c_{2}e^{2t}+7/58 sin(2t)-3/58 cos(2t)