What is the solution for y^{''}-y^'=16e^{3x} ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-y^'=16e^{3x}

y^{^{''}} - y^{^{'}} = 16 e^{3 x}

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

Solution steps

y^{''} (x) - y^{'} (x) = 16 e^{3 x}

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} + 2 y = 5 sin (x) - 10 x e^{x}

y^{^{''}} - 4 y^{^{'}} + 4 y = sin (x)

y^{^{''}} + 49 y = t^{2} cos (7 t) + t e^{- 7 t}

y^{^{'''}} - y^{^{'}} = 11 t

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

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