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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = e^{x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{1}{4} e^{x}

Solution steps

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

Solve linear ODE:

y = e^{x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{1}{4} e^{x}

y = e^{x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{1}{4} e^{x}

y^{^{''}} + 9 y = 2 csc^{2} (3 x)

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

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

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

y^{^{''}} - 3 y^{^{'}} + 2 y = 4 e^{2 t}, y (0) = - 3, y^{^{'}} (0) = 5

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