What is the solution for y^{''}-y^'-2y=4e^{2x} ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} = 2 e^{- t}, y (0) = 2, y^{^{'}} (0) = 2

(D^{2} - 2 D + 1) y = 4 x^{2} e^{x}

x^{^{''}} + 4 x^{^{'}} + 3 x = 4 e^{- 2 t}, x (0) = 0, x^{^{'}} (0) = 0

4 y^{^{''}} - 20 y^{^{'}} + 25 y = 6 - 5 x + 50 x^{2}

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

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