What is the solution for y^{''}-6y^'+8y^'=x^2-1 ?

The solution for y^{''}-6y^'+8y^'=x^2-1 is y=c_{1}+c_{2}e^{-2x}+(x^3)/6-(x^2)/4-x/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-6y^'+8y^'=x^2-1

y^{^{''}} - 6 y^{^{'}} + 8 y^{^{'}} = x^{2} - 1

y = c_{1} + c_{2} e^{- 2 x} + \frac{x ^{3}}{6} - \frac{x ^{2}}{4} - \frac{x}{4}

Solution steps

y^{''} (x) - 6 y^{'} (x) + 8 y^{'} (x) = x^{2} - 1

Solve linear ODE:

y = c_{1} + c_{2} e^{- 2 x} + \frac{x ^{3}}{6} - \frac{x ^{2}}{4} - \frac{x}{4}

y = c_{1} + c_{2} e^{- 2 x} + \frac{x ^{3}}{6} - \frac{x ^{2}}{4} - \frac{x}{4}

y^{^{''}} + 2 y^{^{'}} + y = 4 e^{- 2 t} + t + t^{2}

y^{^{''}} - 16 y = 32 + 45 e^{x}, y (0) = 0, y^{^{''}} (0) = 1

y^{^{''}} - 4 y^{^{'}} - 32 y = - 80 t + 64 t^{2}

D^{3} y = e^{3 x}

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

What is the solution for y^{''}-6y^'+8y^'=x^2-1 ?
The solution for y^{''}-6y^'+8y^'=x^2-1 is y=c_{1}+c_{2}e^{-2x}+(x^3)/6-(x^2)/4-x/4