en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

y = c_{1} e^{2 x} + c_{2} e^{- x} + c_{3} e^{x} + \frac{1}{30} e^{4 x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- x} + c_{3} e^{x} + \frac{1}{30} e^{4 x}

y = c_{1} e^{2 x} + c_{2} e^{- x} + c_{3} e^{x} + \frac{1}{30} e^{4 x}

Popular Examples

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

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

-1/4 y^{''}+y^'+y=x^2-2x

- \frac{1}{4} y^{^{''}} + y^{^{'}} + y = x^{2} - 2 x

y^{''}-2y^'+y=etarctan(t)

y^{^{''}} - 2 y^{^{'}} + y = e t arctan (t)

y^{''}-2y^'+y=(e^t)/((1+t^2))+3e^t

y^{^{''}} - 2 y^{^{'}} + y = \frac{e ^{t}}{( 1 + t ^{2} )} + 3 e^{t}

y^{''}+10y^'+20y=2695e^{2t}cos(11t)

y^{^{''}} + 10 y^{^{'}} + 20 y = 2695 e^{2 t} cos (11 t)

Frequently Asked Questions (FAQ)

What is the solution for y^{'''}-2y^{''}-y^'+2y=e^{4x} ?
The solution for y^{'''}-2y^{''}-y^'+2y=e^{4x} is y=c_{1}e^{2x}+c_{2}e^{-x}+c_{3}e^x+1/30 e^{4x}