What is the solution for 6y^{''}+y^'-y=-25e^{2x}+7e^{-2x} ?

The solution for 6y^{''}+y^'-y=-25e^{2x}+7e^{-2x} is y=c_{1}e^{x/3}+c_{2}e^{-x/2}-e^{2x}+1/3 e^{-2x}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

6y^{''}+y^'-y=-25e^{2x}+7e^{-2x}

6 y^{^{''}} + y^{^{'}} - y = - 25 e^{2 x} + 7 e^{- 2 x}

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

Solution steps

6 y^{''} (x) + y^{'} (x) - y = - 25 e^{2 x} + 7 e^{- 2 x}

Solve linear ODE:

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

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

(d^{3} - 3 d - 2) y = x^{2}

y^{^{''}} - 3 y^{^{'}} - 4 = - 8 e^{t} cos (2 t)

y^{^{''}} - 64 y = - 192 t + 128 t^{2}

y^{^{'''}} - 5 y^{^{''}} + 6 y^{^{'}} = 2 x + 3 e^{2 x}

y^{^{''}} + 2 y^{^{'}} + 2 y = 2 cos (x) + sin (x)

What is the solution for 6y^{''}+y^'-y=-25e^{2x}+7e^{-2x} ?
The solution for 6y^{''}+y^'-y=-25e^{2x}+7e^{-2x} is y=c_{1}e^{x/3}+c_{2}e^{-x/2}-e^{2x}+1/3 e^{-2x}