What is the solution for 2y^{''}+5y^'-3y=0,y(0)=22 ?

The solution for 2y^{''}+5y^'-3y=0,y(0)=22 is y=(22-c_{2})e^{t/2}+c_{2}e^{-3t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2y^{''}+5y^'-3y=0,y(0)=22

2 y^{^{''}} + 5 y^{^{'}} - 3 y = 0, y (0) = 22

y = (22 - c_{2}) e^{\frac{t}{2}} + c_{2} e^{- 3 t}

Solution steps

2 y^{''} (t) + 5 y^{'} (t) - 3 y = 0

Solve linear ODE:

y = (22 - c_{2}) e^{\frac{t}{2}} + c_{2} e^{- 3 t}

y = (22 - c_{2}) e^{\frac{t}{2}} + c_{2} e^{- 3 t}

y^{^{''}} + \frac{1}{r} y^{^{'}} = 0

y^{^{''}} + 0 y^{^{'}} - 9 y = 0

y^{^{''}} = 49 y

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = \frac{7}{4}

\frac{d ^{2} y}{d x ^{2}} + 5 \cdot \frac{d y}{d x} = 0

What is the solution for 2y^{''}+5y^'-3y=0,y(0)=22 ?
The solution for 2y^{''}+5y^'-3y=0,y(0)=22 is y=(22-c_{2})e^{t/2}+c_{2}e^{-3t}