What is the solution for y^{''}+y^'=sin(20x) ?

The solution for y^{''}+y^'=sin(20x) is y=c_{1}+c_{2}e^{-x}-1/401 sin(20x)-1/8020 cos(20x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+y^'=sin(20x)

y^{^{''}} + y^{^{'}} = sin (20 x)

y = c_{1} + c_{2} e^{- x} - \frac{1}{401} sin (20 x) - \frac{1}{8020} cos (20 x)

Solution steps

y^{''} (x) + y^{'} (x) = sin (20 x)

Solve linear ODE:

y = c_{1} + c_{2} e^{- x} - \frac{1}{401} sin (20 x) - \frac{1}{8020} cos (20 x)

y = c_{1} + c_{2} e^{- x} - \frac{1}{401} sin (20 x) - \frac{1}{8020} cos (20 x)

\frac{d y}{d x} = \frac{1}{10} y cos^{2} (y)

\frac{d}{d x} (\frac{x ^{2} + 32 x}{8})

\int_{0}^{1} 2 π (2 - y) (2 y^{2}) d y

\int_{2}^{3} \frac{1}{x ^{2}} d x

\int e^{\frac{x}{6} + 4} d x

What is the solution for y^{''}+y^'=sin(20x) ?
The solution for y^{''}+y^'=sin(20x) is y=c_{1}+c_{2}e^{-x}-1/401 sin(20x)-1/8020 cos(20x)