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

The solution for y^{''}+4y=sin(3x) is y=c_{1}cos(2x)+c_{2}sin(2x)-1/5 sin(3x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{1}{5} sin (3 x)

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{1}{5} sin (3 x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{1}{5} sin (3 x)

f (x) = x^{2} + 1, (2, 5)

\int \frac{2}{1 + 9 x ^{2}} d x

(x^{2} - 1)^{2}

\int x ln (5 + x) d x

\frac{\partial}{\partial y} (3 x^{3} y)

What is the solution for y^{''}+4y=sin(3x) ?
The solution for y^{''}+4y=sin(3x) is y=c_{1}cos(2x)+c_{2}sin(2x)-1/5 sin(3x)