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

The solution for y^{''}+9y=cos(2x) is y=c_{1}cos(3x)+c_{2}sin(3x)+1/5 cos(2x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 10 y^{^{'}} + 25 y = x

d (x) = - 3 x^{2} - 15 x + 146

\frac{d ^{2} y}{d x ^{2}} + y = x sin (x)

\frac{1}{6} y^{^{''}} + 6 y = tan (6 t) - \frac{1}{2} e^{2 t}

9 y^{^{''}} - y = x e^{\frac{x}{3}}, y (0) = 1, y (0) = 0

What is the solution for y^{''}+9y=cos(2x) ?
The solution for y^{''}+9y=cos(2x) is y=c_{1}cos(3x)+c_{2}sin(3x)+1/5 cos(2x)