What is the solution for y^{''}-y^'-12y=5e^{2x}-4 ?

The solution for y^{''}-y^'-12y=5e^{2x}-4 is y=c_{1}e^{4x}+c_{2}e^{-3x}-1/2 e^{2x}+1/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-y^'-12y=5e^{2x}-4

y^{^{''}} - y^{^{'}} - 12 y = 5 e^{2 x} - 4

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

Solution steps

y^{''} (x) - y^{'} (x) - 12 y = 5 e^{2 x} - 4

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} + 8 y = x + e^{x}

4 y^{^{''}} + 11 y^{^{'}} - 3 y = 2 x e^{- 3 x}

y^{^{''}} + 4 y = 2 e^{2 t} + 4 cos (3 t)

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

y^{^{''}} - 16 y^{^{'}} = 2 e^{4 x}

What is the solution for y^{''}-y^'-12y=5e^{2x}-4 ?
The solution for y^{''}-y^'-12y=5e^{2x}-4 is y=c_{1}e^{4x}+c_{2}e^{-3x}-1/2 e^{2x}+1/3