What is the solution for y^{''}-5y^'+6y=te^t ?

The solution for y^{''}-5y^'+6y=te^t is y=c_{1}e^{3t}+c_{2}e^{2t}+(e^tt)/2+3/4 e^t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-5y^'+6y=te^t

y^{^{''}} - 5 y^{^{'}} + 6 y = t e^{t}

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

Solution steps

y^{''} (t) - 5 y^{'} (t) + 6 y = t e^{t}

Solve linear ODE:

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

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

2 y^{^{''}} + 2 y^{^{'}} + y = 4 x

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

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

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

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

What is the solution for y^{''}-5y^'+6y=te^t ?
The solution for y^{''}-5y^'+6y=te^t is y=c_{1}e^{3t}+c_{2}e^{2t}+(e^tt)/2+3/4 e^t