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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-4y^'+4y=x

y^{^{''}} - 4 y^{^{'}} + 4 y = x

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 4 y = t^{2} + 2 e^{t}, y (0) = 0, y^{^{'}} (0) = 1

(d^{4} + 10 d^{2} + 9) y = sin (3 x) + cos (3 x)

y^{^{''}} - 10 y^{^{'}} + 25 y = 30 x + 12

y^{^{''}} + 2 y^{^{'}} - 3 y = 4 e^{x}

y^{^{''}} + 2 y = sin (x)

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