What is the solution for y^{''}+4y=3sin(2x)+x^2-1 ?

The solution for y^{''}+4y=3sin(2x)+x^2-1 is y=c_{1}cos(2x)+c_{2}sin(2x)+(-3xcos(2x)+x^2)/4-3/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{- 3 x cos ( 2 x ) + x ^{2}}{4} - \frac{3}{8}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{- 3 x cos ( 2 x ) + x ^{2}}{4} - \frac{3}{8}

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{- 3 x cos ( 2 x ) + x ^{2}}{4} - \frac{3}{8}

What is the solution for y^{''}+4y=3sin(2x)+x^2-1 ?
The solution for y^{''}+4y=3sin(2x)+x^2-1 is y=c_{1}cos(2x)+c_{2}sin(2x)+(-3xcos(2x)+x^2)/4-3/8