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

The solution for y^{''}-3y^'-4y=(2x-3)e^{2x} is y=c_{1}e^{4x}+c_{2}e^{-x}-(e^{2x}x)/3+4/9 e^{2x}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

y = c_{1} e^{4 x} + c_{2} e^{- x} - \frac{e ^{2 x} x}{3} + \frac{4}{9} e^{2 x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{- x} - \frac{e ^{2 x} x}{3} + \frac{4}{9} e^{2 x}

y = c_{1} e^{4 x} + c_{2} e^{- x} - \frac{e ^{2 x} x}{3} + \frac{4}{9} e^{2 x}

What is the solution for y^{''}-3y^'-4y=(2x-3)e^{2x} ?
The solution for y^{''}-3y^'-4y=(2x-3)e^{2x} is y=c_{1}e^{4x}+c_{2}e^{-x}-(e^{2x}x)/3+4/9 e^{2x}