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

The solution for y^{''}-2y+y=x^2e^x is y=c_{1}e^x+c_{2}e^{-x}+(e^xx^3)/6-(e^xx^2)/4+(e^xx)/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = c_{1} e^{x} + c_{2} e^{- x} + \frac{e ^{x} x ^{3}}{6} - \frac{e ^{x} x ^{2}}{4} + \frac{e ^{x} x}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- x} + \frac{e ^{x} x ^{3}}{6} - \frac{e ^{x} x ^{2}}{4} + \frac{e ^{x} x}{4}

y = c_{1} e^{x} + c_{2} e^{- x} + \frac{e ^{x} x ^{3}}{6} - \frac{e ^{x} x ^{2}}{4} + \frac{e ^{x} x}{4}

\frac{d ^{2} y}{d x ^{2}} - 16 y = 9 e^{4 x} + 30 e^{x}

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

2 y^{^{''}} + 3 y^{^{'}} + y = lo g_{10} (x)

2 y^{^{''}} + 18 y = 6 tan (3 t)

x^{^{''}} - 5 x^{^{'}} - 6 x = 2 e^{- 3 t}

What is the solution for y^{''}-2y+y=x^2e^x ?
The solution for y^{''}-2y+y=x^2e^x is y=c_{1}e^x+c_{2}e^{-x}+(e^xx^3)/6-(e^xx^2)/4+(e^xx)/4