What is the solution for y^{''}+4y^'+4y=5x+4-e^{5x} ?

The solution for y^{''}+4y^'+4y=5x+4-e^{5x} is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+(5x-1)/4-1/49 e^{5x}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{5 x - 1}{4} - \frac{1}{49} e^{5 x}

Solution steps

y^{''} (x) + 4 y^{'} (x) + 4 y = 5 x + 4 - e^{5 x}

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{5 x - 1}{4} - \frac{1}{49} e^{5 x}

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{5 x - 1}{4} - \frac{1}{49} e^{5 x}

y^{^{''}} - 6 y^{^{'}} + 5 y = 34 sin (3 t)

y^{^{''}} - 9 y^{^{'}} + 2 y = t e^{t}

y^{^{''}} + 4 y = 8 sin (2 t) + 5 e^{t}

y^{^{''}} - 8 y^{^{'}} + 16 y = 45 e^{7 t}

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} = 6 x^{2} + 2 x - 2

What is the solution for y^{''}+4y^'+4y=5x+4-e^{5x} ?
The solution for y^{''}+4y^'+4y=5x+4-e^{5x} is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+(5x-1)/4-1/49 e^{5x}