What is the solution for y^{''}-10y^'+26y=4 ?

The solution for y^{''}-10y^'+26y=4 is y=e^{5t}(c_{1}cos(t)+c_{2}sin(t))+2/13

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-10y^'+26y=4

y^{^{''}} - 10 y^{^{'}} + 26 y = 4

y = e^{5 t} (c_{1} cos (t) + c_{2} sin (t)) + \frac{2}{13}

Solution steps

y^{''} (t) - 10 y^{'} (t) + 26 y = 4

Solve linear ODE:

y = e^{5 t} (c_{1} cos (t) + c_{2} sin (t)) + \frac{2}{13}

y = e^{5 t} (c_{1} cos (t) + c_{2} sin (t)) + \frac{2}{13}

y^{^{''}} - 13 y^{^{'}} + 40 y = 27 + 40 x

y^{^{'''}} + 3 y^{^{''}} + 3 y^{^{'}} + 1 = 0

(d^{2} + d - 6) y = e^{- 3 x}

3 y^{^{''}} + 5 y^{^{'}} - y = 10

y^{^{''}} + 2 y^{^{'}} - 8 y = 12 e^{2 x}

What is the solution for y^{''}-10y^'+26y=4 ?
The solution for y^{''}-10y^'+26y=4 is y=e^{5t}(c_{1}cos(t)+c_{2}sin(t))+2/13