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

The solution for y^{''}-8y^'-4y-2=2x is y=c_{1}e^{2(2+sqrt(5))x}+c_{2}e^{2(2-sqrt(5))x}-x/2+1/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = c_{1} e^{2 (2 + 5) x} + c_{2} e^{2 (2 - 5) x} - \frac{x}{2} + \frac{1}{2}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 (2 + 5) x} + c_{2} e^{2 (2 - 5) x} - \frac{x}{2} + \frac{1}{2}

y = c_{1} e^{2 (2 + 5) x} + c_{2} e^{2 (2 - 5) x} - \frac{x}{2} + \frac{1}{2}

y^{^{''}} - 2 y^{^{''}} - 3 y = 4 e^{x} - 9

((d^{2}) - 1) y = x sin (x) + (1 + x^{2}) e^{x}

y^{^{''}} - 5 y^{^{'}} + 6 y = g (x)

y^{^{''}} - 12 y^{^{'}} + 35 y = 96 e^{t}

y^{^{''}} + y + 2 sin (x) = 0

What is the solution for y^{''}-8y^'-4y-2=2x ?
The solution for y^{''}-8y^'-4y-2=2x is y=c_{1}e^{2(2+sqrt(5))x}+c_{2}e^{2(2-sqrt(5))x}-x/2+1/2