What is the solution for y^{''}+9y=2sin(3x) ?

The solution for y^{''}+9y=2sin(3x) is y=c_{1}cos(3x)+c_{2}sin(3x)-(xcos(3x))/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'}} = 4 x y

\int \frac{x ^{2} - x + 1}{( x ^{2} + 1 ) ^{3}} d x

3 e^{x} + \frac{8}{3 x}

\int \frac{6}{1 - x ^{2}} d x

\frac{d}{d x} (ln (π))

What is the solution for y^{''}+9y=2sin(3x) ?
The solution for y^{''}+9y=2sin(3x) is y=c_{1}cos(3x)+c_{2}sin(3x)-(xcos(3x))/3