What is the solution for y^{''}-4y^'+5y=5x^2+2x ?

The solution for y^{''}-4y^'+5y=5x^2+2x is y=e^{2x}(c_{1}cos(x)+c_{2}sin(x))+x^2+2x+6/5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}-4y^'+5y=5x^2+2x

y^{^{''}} - 4 y^{^{'}} + 5 y = 5 x^{2} + 2 x

y = e^{2 x} (c_{1} cos (x) + c_{2} sin (x)) + x^{2} + 2 x + \frac{6}{5}

Solution steps

y^{''} (x) - 4 y^{'} (x) + 5 y = 5 x^{2} + 2 x

Solve linear ODE:

y = e^{2 x} (c_{1} cos (x) + c_{2} sin (x)) + x^{2} + 2 x + \frac{6}{5}

y = e^{2 x} (c_{1} cos (x) + c_{2} sin (x)) + x^{2} + 2 x + \frac{6}{5}

What is the solution for y^{''}-4y^'+5y=5x^2+2x ?
The solution for y^{''}-4y^'+5y=5x^2+2x is y=e^{2x}(c_{1}cos(x)+c_{2}sin(x))+x^2+2x+6/5