What is the solution for y^{''}-5y^'+6y=(2x-1)e^{4x} ?

The solution for y^{''}-5y^'+6y=(2x-1)e^{4x} is y=c_{1}e^{3x}+c_{2}e^{2x}+e^{4x}x-2e^{4x}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-5y^'+6y=(2x-1)e^{4x}

y^{^{''}} - 5 y^{^{'}} + 6 y = (2 x - 1) e^{4 x}

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

Solution steps

y^{''} (x) - 5 y^{'} (x) + 6 y = (2 x - 1) e^{4 x}

Solve linear ODE:

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

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

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

y^{^{''}} + 5 y^{^{'}} + 6 y = - 19 t e^{(2 t)}

y^{^{''}} + 9 y = g (t), y (0) = 4, y^{^{'}} (0) = 0

y^{^{''}} + 5 y^{^{'}} + 4 y = 2 e^{- 5 x}, y (0) = 4, y^{^{'}} (0) = 3

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} - 2 y = 9 x e^{3 x}

What is the solution for y^{''}-5y^'+6y=(2x-1)e^{4x} ?
The solution for y^{''}-5y^'+6y=(2x-1)e^{4x} is y=c_{1}e^{3x}+c_{2}e^{2x}+e^{4x}x-2e^{4x}