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

The solution for y^{''}-y^'+y=x is y=e^{x/2}(c_{1}cos((sqrt(3)x)/2)+c_{2}sin((sqrt(3)x)/2))+x+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = e^{\frac{x}{2}} (c_{1} cos (\frac{3 x}{2}) + c_{2} sin (\frac{3 x}{2})) + x + 1

Solution steps

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

Solve linear ODE:

y = e^{\frac{x}{2}} (c_{1} cos (\frac{3 x}{2}) + c_{2} sin (\frac{3 x}{2})) + x + 1

y = e^{\frac{x}{2}} (c_{1} cos (\frac{3 x}{2}) + c_{2} sin (\frac{3 x}{2})) + x + 1

- y^{^{''}} + y^{^{'}} = 1

y^{^{''''}} - 4 y^{^{''}} = t^{2} - e^{3 t}

y^{^{''}} - 4 y^{^{'}} + 3 y = 4 e^{3 t}, y (0) = 2, y^{^{'}} (0) = 7

\frac{1}{8} y^{^{''}} + 4 y^{^{'}} + 28 x = 0, y (0) = 0, y^{^{'}} (0) = 3

- 6 y^{^{''}} - 5 y^{^{'}} - y = - (t^{2} + 2 t + 2 e^{2 t})

What is the solution for y^{''}-y^'+y=x ?
The solution for y^{''}-y^'+y=x is y=e^{x/2}(c_{1}cos((sqrt(3)x)/2)+c_{2}sin((sqrt(3)x)/2))+x+1