What is the solution for y^{''}+6y^'+25x=0 ?

The solution for y^{''}+6y^'+25x=0 is y=c_{1}+c_{2}e^{-6x}-(25x^2)/(12)+(25x)/(36)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+6y^'+25x=0

y^{^{''}} + 6 y^{^{'}} + 25 x = 0

y = c_{1} + c_{2} e^{- 6 x} - \frac{25 x ^{2}}{12} + \frac{25 x}{36}

Solution steps

y^{''} (x) + 6 y^{'} (x) + 25 x = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- 6 x} - \frac{25 x ^{2}}{12} + \frac{25 x}{36}

y = c_{1} + c_{2} e^{- 6 x} - \frac{25 x ^{2}}{12} + \frac{25 x}{36}

y^{^{''}} - n^{2} y = \frac{2 cos ( πn ) x}{n}

y^{^{''}} + 8 y^{^{'}} - 2 y = \frac{5}{e ^{3 t}}

y^{^{''}} + 4 y = sin (x), y (0) = 0, y (1) = 0

(d - 4)^{6} y = 5 e^{4} x

y^{^{''}} - 81 y = - 243 t + 162 t^{2}

What is the solution for y^{''}+6y^'+25x=0 ?
The solution for y^{''}+6y^'+25x=0 is y=c_{1}+c_{2}e^{-6x}-(25x^2)/(12)+(25x)/(36)