What is the solution for y^{''}-2y^'+y=4x^2-7+x^{-1}e^x ?

The solution for y^{''}-2y^'+y=4x^2-7+x^{-1}e^x is y=c_{1}e^x+c_{2}xe^x+4x^2+16x-e^xx+e^xxln(x)+17

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-2y^'+y=4x^2-7+x^{-1}e^x

y^{^{''}} - 2 y^{^{'}} + y = 4 x^{2} - 7 + x^{- 1} e^{x}

y = c_{1} e^{x} + c_{2} x e^{x} + 4 x^{2} + 16 x - e^{x} x + e^{x} x ln (x) + 17

Solution steps

y^{''} (x) - 2 y^{'} (x) + y = 4 x^{2} - 7 + x^{- 1} e^{x}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + 4 x^{2} + 16 x - e^{x} x + e^{x} x ln (x) + 17

y = c_{1} e^{x} + c_{2} x e^{x} + 4 x^{2} + 16 x - e^{x} x + e^{x} x ln (x) + 17

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

y^{^{''}} - 4 y^{^{'}} + 4 y = 12 x^{2} - 40 x + 42

y^{^{''}} + 6 y^{^{'}} + 73 y = 4 cos (5 x)

y^{^{''}} - 4 y^{^{'}} - 5 = 0

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

What is the solution for y^{''}-2y^'+y=4x^2-7+x^{-1}e^x ?
The solution for y^{''}-2y^'+y=4x^2-7+x^{-1}e^x is y=c_{1}e^x+c_{2}xe^x+4x^2+16x-e^xx+e^xxln(x)+17