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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Solution steps

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

Solve linear ODE:

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

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

d er i v a t i v e \frac{9}{x + 2}

\frac{\partial}{\partial x} (x y e^{z} + x - z^{2} - 3)

\int sin^{3} (10 x) d x

t an g e n t y = x^{2} - x + 4, at x = 4

x \to - 0 lim (\frac{1}{1 + e ^{\frac{1}{x}}})

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