What is the solution for y^{''}-2y^'-8y=xe^x+4 ?

The solution for y^{''}-2y^'-8y=xe^x+4 is y=c_{1}e^{4x}+c_{2}e^{-2x}-(e^xx)/9-1/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-2y^'-8y=xe^x+4

y^{^{''}} - 2 y^{^{'}} - 8 y = x e^{x} + 4

y = c_{1} e^{4 x} + c_{2} e^{- 2 x} - \frac{e ^{x} x}{9} - \frac{1}{2}

Solution steps

y^{''} (x) - 2 y^{'} (x) - 8 y = x e^{x} + 4

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{- 2 x} - \frac{e ^{x} x}{9} - \frac{1}{2}

y = c_{1} e^{4 x} + c_{2} e^{- 2 x} - \frac{e ^{x} x}{9} - \frac{1}{2}

x \to \frac{π}{2} - lim (cos (x))

\frac{d y}{d x} = \frac{( 5 y ^{3} - 2 x y )}{x ^{2}}

(1 + cos^{2} (x))^{4}

y = x^{3} - 16 x^{2} + 45 x + 6

\int 4 sin (x) + 4 d x

What is the solution for y^{''}-2y^'-8y=xe^x+4 ?
The solution for y^{''}-2y^'-8y=xe^x+4 is y=c_{1}e^{4x}+c_{2}e^{-2x}-(e^xx)/9-1/2