What is the solution for linear y^{''}-y=x,y(0)=0,y(1)=0 ?

The solution for linear y^{''}-y=x,y(0)=0,y(1)=0 is y=-e/(-e^2+1)e^x+e/(-e^2+1)e^{-x}-x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

linear y^{''}-y=x,y(0)=0,y(1)=0

linear

y^{^{''}} - y = x, y (0) = 0, y (1) = 0

y = - \frac{e}{- e ^{2} + 1} e^{x} + \frac{e}{- e ^{2} + 1} e^{- x} - x

Solution steps

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

Solve linear ODE:

y = - \frac{e}{- e ^{2} + 1} e^{x} + \frac{e}{- e ^{2} + 1} e^{- x} - x

y = - \frac{e}{- e ^{2} + 1} e^{x} + \frac{e}{- e ^{2} + 1} e^{- x} - x

y^{^{''}} + 2 y^{^{'}} + 2 y = x + 1

y^{^{''}} - 2 y = sin (t)

y^{^{''}} + 2 y^{^{'}} + 5 y = 34 cos (2 x)

(D^{2} - 7 D + 6) y = e^{6 x} + e^{x}

\frac{d ^{2} y}{d x ^{2}} - y = 4 e^{x} cos (x)

What is the solution for linear y^{''}-y=x,y(0)=0,y(1)=0 ?
The solution for linear y^{''}-y=x,y(0)=0,y(1)=0 is y=-e/(-e^2+1)e^x+e/(-e^2+1)e^{-x}-x