What is the solution for y^{''}-4y^'+4y=e^{-x}(9x^2-3x-4) ?

The solution for y^{''}-4y^'+4y=e^{-x}(9x^2-3x-4) is y=c_{1}e^{2x}+c_{2}xe^{2x}+e^{-x}x^2+e^{-x}x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-4y^'+4y=e^{-x}(9x^2-3x-4)

y^{^{''}} - 4 y^{^{'}} + 4 y = e^{- x} (9 x^{2} - 3 x - 4)

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

Solution steps

y^{''} (x) - 4 y^{'} (x) + 4 y = e^{- x} (9 x^{2} - 3 x - 4)

Solve linear ODE:

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

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

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

y^{^{''}} + 4 y^{^{'}} + 3 y = 64 e^{t}, y (0) = 4, y^{^{'}} (0) = 7

x^{^{''}} + x^{^{'}} + 2 x = sin (t)

q^{^{''}} + 10 q^{^{'}} + 10000 q = 20 (t)

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

What is the solution for y^{''}-4y^'+4y=e^{-x}(9x^2-3x-4) ?
The solution for y^{''}-4y^'+4y=e^{-x}(9x^2-3x-4) is y=c_{1}e^{2x}+c_{2}xe^{2x}+e^{-x}x^2+e^{-x}x