What is the solution for y^{''}+2y^'-3y=-16xe^x ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+2y^'-3y=-16xe^x

y^{^{''}} + 2 y^{^{'}} - 3 y = - 16 x e^{x}

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

Solution steps

y^{''} (x) + 2 y^{'} (x) - 3 y = - 16 x e^{x}

Solve linear ODE:

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

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

y = 3 - x, y = 2 x^{2}, [0, 3]

\frac{d}{d x} (4 x + 2 y)

\frac{\partial}{\partial x} (- 6 x^{3} + y^{2} + 5 x y)

\frac{- 2 x}{( x ^{2} + 1 ) ^{2}}

\frac{1}{x} + e^{x}

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