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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

2 y^{^{''}} + 3 y^{^{'}} + y = lo g_{10} (x)

2 y^{^{''}} + 18 y = 6 tan (3 t)

x^{^{''}} - 5 x^{^{'}} - 6 x = 2 e^{- 3 t}

3 y^{^{''}} + 2 g^{2} y = 4.9 g cos (8 t)

x^{^{''}} + 2.8 x^{^{'}} + 4 x = 10, x (0) = 2, x^{^{'}} (0) = 3

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