What is the solution for y^{''}+5y^'-6y=(3x-1)e^x ?

The solution for y^{''}+5y^'-6y=(3x-1)e^x is y=c_{1}e^x+c_{2}e^{-6x}+(3e^xx^2)/(14)-(10e^xx)/(49)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+5y^'-6y=(3x-1)e^x

y^{^{''}} + 5 y^{^{'}} - 6 y = (3 x - 1) e^{x}

y = c_{1} e^{x} + c_{2} e^{- 6 x} + \frac{3 e ^{x} x ^{2}}{14} - \frac{10 e ^{x} x}{49}

Solution steps

y^{''} (x) + 5 y^{'} (x) - 6 y = (3 x - 1) e^{x}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 6 x} + \frac{3 e ^{x} x ^{2}}{14} - \frac{10 e ^{x} x}{49}

y = c_{1} e^{x} + c_{2} e^{- 6 x} + \frac{3 e ^{x} x ^{2}}{14} - \frac{10 e ^{x} x}{49}

e^{x^{4}}

x + y = 7

\frac{d}{d x} (x^{c})

\frac{d}{d x} (3 x^{\frac{4}{3}} - \frac{3 x ^{\frac{2}{3}}}{32})

\int t^{2} cos^{2} (t) d t

What is the solution for y^{''}+5y^'-6y=(3x-1)e^x ?
The solution for y^{''}+5y^'-6y=(3x-1)e^x is y=c_{1}e^x+c_{2}e^{-6x}+(3e^xx^2)/(14)-(10e^xx)/(49)