What is the solution for 4y^{''}-4y^'+y=(4x)/(e^{x/2)} ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} + y = \frac{1}{6} e^{- x}

y^{^{''}} - 6 y^{^{'}} + 9 y = 12 e^{x} sin (x)

y^{^{''}} + 6 y^{^{'}} + 9 y = \frac{e ^{- 3 x}}{x ^{2}}

2 y^{^{''}} + y^{^{'}} = 4 x, y (0) = 7, y^{^{'}} (0) = - 10

y^{^{''}} + 4 y = r (x)

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