What is the solution for y^{''}-y^'=e^{-3t},y(0)=0,y^'(0)=0 ?

The solution for y^{''}-y^'=e^{-3t},y(0)=0,y^'(0)=0 is y=(1/3-1/12)e^t-1/3+1/12 e^{-3t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-y^'=e^{-3t},y(0)=0,y^'(0)=0

y^{^{''}} - y^{^{'}} = e^{- 3 t}, y (0) = 0, y^{^{'}} (0) = 0

y = (\frac{1}{3} - \frac{1}{12}) e^{t} - \frac{1}{3} + \frac{1}{12} e^{- 3 t}

Solution steps

y^{''} (t) - y^{'} (t) = e^{- 3 t}

Solve linear ODE:

y = (\frac{1}{3} - \frac{1}{12}) e^{t} - \frac{1}{3} + \frac{1}{12} e^{- 3 t}

y = (\frac{1}{3} - \frac{1}{12}) e^{t} - \frac{1}{3} + \frac{1}{12} e^{- 3 t}

y^{^{''}} + 8 y^{^{'}} + 16 = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} - 2 y^{^{'}} = 6 x + 6 - 2 e^{4 x}

y^{^{''}} - 4 y^{^{'}} + 4 y = 15 sin (x) - 35 cos (x)

y^{^{''}} + 9 y = cos (2 t), y (\frac{π}{2}) = - 1, y (0) = 1

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

What is the solution for y^{''}-y^'=e^{-3t},y(0)=0,y^'(0)=0 ?
The solution for y^{''}-y^'=e^{-3t},y(0)=0,y^'(0)=0 is y=(1/3-1/12)e^t-1/3+1/12 e^{-3t}