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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-2y^'+y=x+2+2cos(x)

y^{^{''}} - 2 y^{^{'}} + y = x + 2 + 2 cos (x)

y = c_{1} e^{x} + c_{2} x e^{x} + x + 4 - sin (x)

Solution steps

y^{''} (x) - 2 y^{'} (x) + y = x + 2 + 2 cos (x)

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + x + 4 - sin (x)

y = c_{1} e^{x} + c_{2} x e^{x} + x + 4 - sin (x)

(D^{3} + 10 D^{2} + 25 D) y = e^{x}

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

y^{^{''}} + 3 y^{^{'}} = 4

2 y^{^{''}} + y^{^{'}} - y = 3 x + 4, y (0) = 1, y^{^{'}} (0) = 1

y^{^{''}} + 9 y = 18 sin (3 t)

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