What is the solution for y^{''}-5y^'+6y=x^2-3x+1 ?

The solution for y^{''}-5y^'+6y=x^2-3x+1 is y=c_{1}e^{3x}+c_{2}e^{2x}+(x^2)/6-(2x)/9-2/27

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-5y^'+6y=x^2-3x+1

y^{^{''}} - 5 y^{^{'}} + 6 y = x^{2} - 3 x + 1

y = c_{1} e^{3 x} + c_{2} e^{2 x} + \frac{x ^{2}}{6} - \frac{2 x}{9} - \frac{2}{27}

Solution steps

y^{''} (x) - 5 y^{'} (x) + 6 y = x^{2} - 3 x + 1

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{2 x} + \frac{x ^{2}}{6} - \frac{2 x}{9} - \frac{2}{27}

y = c_{1} e^{3 x} + c_{2} e^{2 x} + \frac{x ^{2}}{6} - \frac{2 x}{9} - \frac{2}{27}

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

y^{^{''}} + 2 y^{^{'}} + 2 y = e^{- 1}

2 y^{^{''}} + 3 y^{^{'}} - 2 y = - 8 x - 13, y (0) = 0, y^{^{'}} (0) = 0

2 x^{^{''}} - x^{^{'}} + 4 x = cos (3 t)

y^{^{''}} + 9 y = 5 sin (x)

What is the solution for y^{''}-5y^'+6y=x^2-3x+1 ?
The solution for y^{''}-5y^'+6y=x^2-3x+1 is y=c_{1}e^{3x}+c_{2}e^{2x}+(x^2)/6-(2x)/9-2/27