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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-y=sin^2(x)

y^{^{''}} - y = sin^{2} (x)

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

Solution steps

y^{''} (x) - y = sin^{2} (x)

Solve linear ODE:

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

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

(D^{2} + 4) y = 8 cos^{2} (x)

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

y^{^{''}} + 4 y^{^{'}} + 4 y = e^{- x} cos (x)

y^{^{''}} + 3 y^{^{'}} - 4 y = e^{x}

y^{^{'''}} + 2 y^{^{''}} + y^{^{'}} = 10

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