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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

linear y^{''}+y=3cos(2x)

linear

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y^{^{'}} = \frac{1}{sin ( t )}

\frac{d ^{2} x}{d t ^{2}} + 5 \frac{d x}{d t} + 3 x = t + 1

y^{^{''}} - 9 y^{^{'}} = x + 1

y^{^{''}} - 2 y^{^{'}} + y = 4 x e^{x}

y^{^{''}} + 3 y^{^{'}} = 10

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