What is the solution for y^{''}-y=t,y(0)=0,y^'(0)=1 ?

The solution for y^{''}-y=t,y(0)=0,y^'(0)=1 is y=e^t-e^{-t}-t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-y=t,y(0)=0,y^'(0)=1

y^{^{''}} - y = t, y (0) = 0, y^{^{'}} (0) = 1

y = e^{t} - e^{- t} - t

Solution steps

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

Solve linear ODE:

y = e^{t} - e^{- t} - t

y = e^{t} - e^{- t} - t

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

y^{^{''}} + y^{^{'}} = 2 x + 3

y^{^{''}} - 4 y^{^{'}} + 4 y = 14 x + 5

y^{^{''}} + 10 y^{^{'}} + 25 y = e^{\frac{- 10 x}{x ^{2}}}

d (r) = \frac{( ln ( 2 ) )}{ln ( 1 + 0.04 )}

What is the solution for y^{''}-y=t,y(0)=0,y^'(0)=1 ?
The solution for y^{''}-y=t,y(0)=0,y^'(0)=1 is y=e^t-e^{-t}-t