What is the solution for y^{''}+3y^'-10y=0 ?

The solution for y^{''}+3y^'-10y=0 is y=c_{1}e^{2t}+c_{2}e^{-5t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+3y^'-10y=0

y^{^{''}} + 3 y^{^{'}} - 10 y = 0

y = c_{1} e^{2 t} + c_{2} e^{- 5 t}

Solution steps

y^{''} (t) + 3 y^{'} (t) - 10 y = 0

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- 5 t}

y = c_{1} e^{2 t} + c_{2} e^{- 5 t}

\frac{d}{d x} (e^{x^{4}} \cdot 4 x^{3})

x^{^{''}} + k x = 0

y = \frac{4}{x}, y = 4 x, x = 4, x = 0

\frac{d y}{d x} = 10 - \frac{2 y}{100 - x}

\frac{d}{d x} (e^{x} + 3 e^{- x})

What is the solution for y^{''}+3y^'-10y=0 ?
The solution for y^{''}+3y^'-10y=0 is y=c_{1}e^{2t}+c_{2}e^{-5t}