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

The solution for y^{''}+2y^'-3y=(x+1)e^x is y=c_{1}e^x+c_{2}e^{-3x}+(e^xx^2)/8+(3e^xx)/(16)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+2y^'-3y=(x+1)e^x

y^{^{''}} + 2 y^{^{'}} - 3 y = (x + 1) e^{x}

y = c_{1} e^{x} + c_{2} e^{- 3 x} + \frac{e ^{x} x ^{2}}{8} + \frac{3 e ^{x} x}{16}

Solution steps

y^{''} (x) + 2 y^{'} (x) - 3 y = (x + 1) e^{x}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 3 x} + \frac{e ^{x} x ^{2}}{8} + \frac{3 e ^{x} x}{16}

y = c_{1} e^{x} + c_{2} e^{- 3 x} + \frac{e ^{x} x ^{2}}{8} + \frac{3 e ^{x} x}{16}

\frac{d ^{2} y}{d x ^{2}} - 14 \frac{d y}{d x} + 49 y = (x + 1) e^{x}

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

6 y^{^{''}} - 5 y^{^{'}} + y = 3 x^{2} + x

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 6 y = e^{2 x}

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

What is the solution for y^{''}+2y^'-3y=(x+1)e^x ?
The solution for y^{''}+2y^'-3y=(x+1)e^x is y=c_{1}e^x+c_{2}e^{-3x}+(e^xx^2)/8+(3e^xx)/(16)