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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} - 5 y^{^{'}} + 6 y = e^{- 5 x} (95 + 250 x + 56 x^{2})

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 4 y = 1

y^{^{''}} + 2 y^{^{'}} + y = 4 sin (x)

y^{^{''}} - 2 y - 24 = 0

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