What is the solution for 2y^{''}-4y^'+10y=2x ?

The solution for 2y^{''}-4y^'+10y=2x is y=e^x(c_{1}cos(2x)+c_{2}sin(2x))+x/5+2/25

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2y^{''}-4y^'+10y=2x

2 y^{^{''}} - 4 y^{^{'}} + 10 y = 2 x

y = e^{x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{x}{5} + \frac{2}{25}

Solution steps

2 y^{''} (x) - 4 y^{'} (x) + 10 y = 2 x

Solve linear ODE:

y = e^{x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{x}{5} + \frac{2}{25}

y = e^{x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{x}{5} + \frac{2}{25}

0.25 y^{^{''}} + y^{^{'}} + y = x^{2} - 2 x

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} + y = x

y^{^{''}} - 6 y^{^{'}} + 9 y = sin (2 x) + x

\frac{1}{2} x^{^{''}} + \frac{1}{2} x^{^{'}} + 6 x = 20 cos (3 t)

y^{^{''}} + 4 y^{^{'}} + 6 y = 4 e^{- 2 t} cos (t)

What is the solution for 2y^{''}-4y^'+10y=2x ?
The solution for 2y^{''}-4y^'+10y=2x is y=e^x(c_{1}cos(2x)+c_{2}sin(2x))+x/5+2/25