What is the solution for y^{''}-y^'+5y=3 ?

The solution for y^{''}-y^'+5y=3 is y=e^{t/2}(c_{1}cos((sqrt(19)t)/2)+c_{2}sin((sqrt(19)t)/2))+3/5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-y^'+5y=3

y^{^{''}} - y^{^{'}} + 5 y = 3

y = e^{\frac{t}{2}} (c_{1} cos (\frac{19 t}{2}) + c_{2} sin (\frac{19 t}{2})) + \frac{3}{5}

Solution steps

y^{''} (t) - y^{'} (t) + 5 y = 3

Solve linear ODE:

y = e^{\frac{t}{2}} (c_{1} cos (\frac{19 t}{2}) + c_{2} sin (\frac{19 t}{2})) + \frac{3}{5}

y = e^{\frac{t}{2}} (c_{1} cos (\frac{19 t}{2}) + c_{2} sin (\frac{19 t}{2})) + \frac{3}{5}

\frac{d}{d x} (2 (1 - x))

y^{^{''}} + 4 y = x^{2} + 3 e^{x}

\frac{\partial}{\partial x} (cos (k x))

(- 4, 3), (- 1, - 6)

\int x^{2} 4 - x^{2} d x

What is the solution for y^{''}-y^'+5y=3 ?
The solution for y^{''}-y^'+5y=3 is y=e^{t/2}(c_{1}cos((sqrt(19)t)/2)+c_{2}sin((sqrt(19)t)/2))+3/5