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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-2y^'-3y=e^{-x}

y^{^{''}} - 2 y^{^{'}} - 3 y = e^{- x}

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

Solution steps

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

Solve linear ODE:

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

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

sin^{2} (5 x)

d y + y d x = 2 x y^{2} e^{x} d x

\frac{d}{d x} (2^{x} x^{2})

\int \frac{1}{8 - 7 x} d x

\int (x^{2} + 2 x) cos (x) d x

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