What is the solution for y^{''}+4y^'+4y=15x+17 ?

The solution for y^{''}+4y^'+4y=15x+17 is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+(15x)/4+1/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y^'+4y=15x+17

y^{^{''}} + 4 y^{^{'}} + 4 y = 15 x + 17

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{15 x}{4} + \frac{1}{2}

Solution steps

y^{''} (x) + 4 y^{'} (x) + 4 y = 15 x + 17

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{15 x}{4} + \frac{1}{2}

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{15 x}{4} + \frac{1}{2}

y^{^{''}} + 16 y^{^{'}} + 64 y = 100 e^{2 x}

2 y^{^{''}} - 5 y^{^{'}} + 71.86 y = 4

y^{^{''}} - 2 y^{^{'}} - 3 y = 6 x e^{x} + 4 x - 5

y^{^{''}} - 2 y^{^{'}} + 5 y = cos (x)

y^{^{''}} + 3 y = 5 t^{4}, y (0) = 0, y^{^{'}} (0) = 0

What is the solution for y^{''}+4y^'+4y=15x+17 ?
The solution for y^{''}+4y^'+4y=15x+17 is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+(15x)/4+1/2