What is the solution for y^{''}+8y^'+16y=x^2 ?

The solution for y^{''}+8y^'+16y=x^2 is y=c_{1}e^{-4x}+c_{2}xe^{-4x}+(x^2)/(16)-x/(16)+3/128

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+8y^'+16y=x^2

y^{^{''}} + 8 y^{^{'}} + 16 y = x^{2}

y = c_{1} e^{- 4 x} + c_{2} x e^{- 4 x} + \frac{x ^{2}}{16} - \frac{x}{16} + \frac{3}{128}

Solution steps

y^{''} (x) + 8 y^{'} (x) + 16 y = x^{2}

Solve linear ODE:

y = c_{1} e^{- 4 x} + c_{2} x e^{- 4 x} + \frac{x ^{2}}{16} - \frac{x}{16} + \frac{3}{128}

y = c_{1} e^{- 4 x} + c_{2} x e^{- 4 x} + \frac{x ^{2}}{16} - \frac{x}{16} + \frac{3}{128}

(d^{3} - 3 d + 2) \cdot y = x

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

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{3} t

(d^{2} - d - 2) y = 6 e^{\frac{- t}{2}}

d, d (q) = - 0.4 q + 23

What is the solution for y^{''}+8y^'+16y=x^2 ?
The solution for y^{''}+8y^'+16y=x^2 is y=c_{1}e^{-4x}+c_{2}xe^{-4x}+(x^2)/(16)-x/(16)+3/128