What is the solution for 2y^{''}+5y^'-3y=0,y(0)=-7,y^'(0)=28 ?

The solution for 2y^{''}+5y^'-3y=0,y(0)=-7,y^'(0)=28 is y=2e^{t/2}-9e^{-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)=-7,y^'(0)=28

2 y^{^{''}} + 5 y^{^{'}} - 3 y = 0, y (0) = - 7, y^{^{'}} (0) = 28

y = 2 e^{\frac{t}{2}} - 9 e^{- 3 t}

Solution steps

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

Solve linear ODE:

y = 2 e^{\frac{t}{2}} - 9 e^{- 3 t}

y = 2 e^{\frac{t}{2}} - 9 e^{- 3 t}

(2 e^{2 x})^{^{'}}

- 9 (6 x^{2} + 7)^{- 6}

(x^{2} + y^{2})^{2}

\frac{\partial}{\partial y} (3 z e^{x yz})

\frac{x ^{2}}{x + 8}

What is the solution for 2y^{''}+5y^'-3y=0,y(0)=-7,y^'(0)=28 ?
The solution for 2y^{''}+5y^'-3y=0,y(0)=-7,y^'(0)=28 is y=2e^{t/2}-9e^{-3t}