What is the solution for y^{''}+2y^'-3y=t,y(0)=2,y^'(0)=1 ?

The solution for y^{''}+2y^'-3y=t,y(0)=2,y^'(0)=1 is y=(-2/9+20/9)e^t+2/9 e^{-3t}-t/3-2/9

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+2y^'-3y=t,y(0)=2,y^'(0)=1

y^{^{''}} + 2 y^{^{'}} - 3 y = t, y (0) = 2, y^{^{'}} (0) = 1

y = (- \frac{2}{9} + \frac{20}{9}) e^{t} + \frac{2}{9} e^{- 3 t} - \frac{t}{3} - \frac{2}{9}

Solution steps

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

Solve linear ODE:

y = (- \frac{2}{9} + \frac{20}{9}) e^{t} + \frac{2}{9} e^{- 3 t} - \frac{t}{3} - \frac{2}{9}

y = (- \frac{2}{9} + \frac{20}{9}) e^{t} + \frac{2}{9} e^{- 3 t} - \frac{t}{3} - \frac{2}{9}

y^{^{''}} + 4 y = 3 sin (x) + 3 cos (x)

y^{^{''}} - 2 y^{^{'}} = 8

y^{^{''}} - 2 y^{^{'}} = 1

y^{^{''}} - 2 y^{^{'}} = 4

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

What is the solution for y^{''}+2y^'-3y=t,y(0)=2,y^'(0)=1 ?
The solution for y^{''}+2y^'-3y=t,y(0)=2,y^'(0)=1 is y=(-2/9+20/9)e^t+2/9 e^{-3t}-t/3-2/9