What is the solution for 2y^{''}-8y=16e^x ?

The solution for 2y^{''}-8y=16e^x is y=c_{1}e^{2x}+c_{2}e^{-2x}-8/3 e^x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2y^{''}-8y=16e^x

2 y^{^{''}} - 8 y = 16 e^{x}

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - \frac{8}{3} e^{x}

Solution steps

2 y^{''} (x) - 8 y = 16 e^{x}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - \frac{8}{3} e^{x}

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - \frac{8}{3} e^{x}

(d^{3} - 3 d - 2) y = 540 x^{3} e^{- x}

y^{^{''}} + y = 2 sin (2 t), y (0) = 9, y^{^{'}} (0) = 0

y^{^{''}} + 4 y = sec (x)

y^{^{''}} + 2 y^{^{'}} + y = \frac{( e ^{x} )}{x}

y^{^{''}} + 2 y^{^{'}} + y = cos (5 t)

What is the solution for 2y^{''}-8y=16e^x ?
The solution for 2y^{''}-8y=16e^x is y=c_{1}e^{2x}+c_{2}e^{-2x}-8/3 e^x