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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = c_{1} e^{2 x} + c_{2} + \frac{- x ^{2} - x}{4} - \frac{1}{5} sin (x) + \frac{2}{5} cos (x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} + \frac{- x ^{2} - x}{4} - \frac{1}{5} sin (x) + \frac{2}{5} cos (x)

y = c_{1} e^{2 x} + c_{2} + \frac{- x ^{2} - x}{4} - \frac{1}{5} sin (x) + \frac{2}{5} cos (x)

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

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

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

y^{^{''''}} + 49 y^{^{''}} = sin (7 t) + t e^{8 t} + 4

y^{^{''}} = y^{^{'}} + x

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