What is the solution for 3y^{''}+y^'=9x ?

The solution for 3y^{''}+y^'=9x is y=c_{1}+c_{2}e^{-x/3}+(9x^2)/2-27x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3y^{''}+y^'=9x

3 y^{^{''}} + y^{^{'}} = 9 x

y = c_{1} + c_{2} e^{- \frac{x}{3}} + \frac{9 x ^{2}}{2} - 27 x

Solution steps

3 y^{''} (x) + y^{'} (x) = 9 x

Solve linear ODE:

y = c_{1} + c_{2} e^{- \frac{x}{3}} + \frac{9 x ^{2}}{2} - 27 x

y = c_{1} + c_{2} e^{- \frac{x}{3}} + \frac{9 x ^{2}}{2} - 27 x

x^{^{''}} + 5 x^{^{'}} + 6 x = 2 e^{- t}, x (0) = 1, x^{^{'}} (0) = 0

y^{^{''}} - y = 2 sin (x)

x^{^{''}} - x = 4 e^{t}

y^{^{''}} + y^{^{'}} - 2 y = x^{2} + 3

y^{^{'''}} + y^{^{'}} = e^{x}, y (0) = 0

What is the solution for 3y^{''}+y^'=9x ?
The solution for 3y^{''}+y^'=9x is y=c_{1}+c_{2}e^{-x/3}+(9x^2)/2-27x