What is the solution for y^{''}+y^'=7,y(0)=1,y^'(0)=0 ?

The solution for y^{''}+y^'=7,y(0)=1,y^'(0)=0 is y=7t+7e^{-t}-6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+y^'=7,y(0)=1,y^'(0)=0

y^{^{''}} + y^{^{'}} = 7, y (0) = 1, y^{^{'}} (0) = 0

y = 7 t + 7 e^{- t} - 6

Solution steps

y^{''} (t) + y^{'} (t) = 7

Solve linear ODE:

y = 7 t + 7 e^{- t} - 6

y = 7 t + 7 e^{- t} - 6

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

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

y^{^{''''}} - y^{^{''}} = 2 sin (t) - 3 e^{- t} + 4 t

y^{^{''}} + 4 y = cos (4 t) + 2 sin (4 t)

y^{^{''}} - 4 y = 2 x - 3 cos (x) + e^{- x}

What is the solution for y^{''}+y^'=7,y(0)=1,y^'(0)=0 ?
The solution for y^{''}+y^'=7,y(0)=1,y^'(0)=0 is y=7t+7e^{-t}-6