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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

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

y, (d^{3} + d) \cdot y = cos (t) + t^{2} + 3

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

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

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