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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{\frac{x}{2}} x + \frac{e ^{\frac{x}{2}} x ^{2}}{8}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{\frac{x}{2}} x + \frac{e ^{\frac{x}{2}} x ^{2}}{8}

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{\frac{x}{2}} x + \frac{e ^{\frac{x}{2}} x ^{2}}{8}

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

y^{^{''}} - 4 y^{^{'}} + 13 (t) = 0

y^{^{''}} - y^{^{'}} = 2 e^{- t}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 9 y = 12 cos (3 t)

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

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