What is the solution for y^{''}-2y^'+2y=e^tsin(t) ?

The solution for y^{''}-2y^'+2y=e^tsin(t) is y=e^t(c_{1}cos(t)+c_{2}sin(t))-(e^ttcos(t))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}-2y^'+2y=e^tsin(t)

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{t} sin (t)

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

Solution steps

y^{''} (t) - 2 y^{'} (t) + 2 y = e^{t} sin (t)

Solve linear ODE:

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

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

What is the solution for y^{''}-2y^'+2y=e^tsin(t) ?
The solution for y^{''}-2y^'+2y=e^tsin(t) is y=e^t(c_{1}cos(t)+c_{2}sin(t))-(e^ttcos(t))/2