What is the solution for linear y^{''}-2y^'+5y=6e^t,y(0)=0 ?

The solution for linear y^{''}-2y^'+5y=6e^t,y(0)=0 is y=e^t(-3/2 cos(2t)+c_{2}sin(2t))+3/2 e^t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

linear y^{''}-2y^'+5y=6e^t,y(0)=0

linear

y^{^{''}} - 2 y^{^{'}} + 5 y = 6 e^{t}, y (0) = 0

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

Solution steps

y^{''} (t) - 2 y^{'} (t) + 5 y = 6 e^{t}

Solve linear ODE:

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

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

What is the solution for linear y^{''}-2y^'+5y=6e^t,y(0)=0 ?
The solution for linear y^{''}-2y^'+5y=6e^t,y(0)=0 is y=e^t(-3/2 cos(2t)+c_{2}sin(2t))+3/2 e^t