What is the solution for y^{''}-5y^'+6y=2et ?

The solution for y^{''}-5y^'+6y=2et is y=c_{1}e^{3t}+c_{2}e^{2t}+(et)/3+(5e)/(18)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-5y^'+6y=2et

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

y = c_{1} e^{3 t} + c_{2} e^{2 t} + \frac{e t}{3} + \frac{5 e}{18}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} e^{2 t} + \frac{e t}{3} + \frac{5 e}{18}

y = c_{1} e^{3 t} + c_{2} e^{2 t} + \frac{e t}{3} + \frac{5 e}{18}

y^{^{''}} + 2 y^{^{'}} + y = 2 cos (2 x) + 3 x + 2 + e^{x}

y^{^{''}} - 2 y^{^{'}} + 17 y = 289 x + 9

y^{^{''}} + y^{^{'}} = (2 t)

y^{^{''}} + 3 y + 2 = 0, y (2) = - 1, y^{^{'}} (0) = 0

y^{^{''}} - 2 y^{^{'}} + 2 y = e - t, y (0) = 0, y^{^{'}} (0) = 1

What is the solution for y^{''}-5y^'+6y=2et ?
The solution for y^{''}-5y^'+6y=2et is y=c_{1}e^{3t}+c_{2}e^{2t}+(et)/3+(5e)/(18)