What is the solution for y^{''}-3y^'-4y=0 ?

The solution for y^{''}-3y^'-4y=0 is y=c_{1}e^{4t}+c_{2}e^{-t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-3y^'-4y=0

y^{^{''}} - 3 y^{^{'}} - 4 y = 0

y = c_{1} e^{4 t} + c_{2} e^{- t}

Solution steps

y^{''} (t) - 3 y^{'} (t) - 4 y = 0

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} e^{- t}

y = c_{1} e^{4 t} + c_{2} e^{- t}

n = 1 \sum \infty \frac{1}{n + 8}

\frac{d}{d x} ((x + 3) (2 x - 1)^{2})

y = cos^{4} (x)

sin (x), (\frac{π}{2})

\frac{d}{d x} (x^{3} - 3 x^{2})

What is the solution for y^{''}-3y^'-4y=0 ?
The solution for y^{''}-3y^'-4y=0 is y=c_{1}e^{4t}+c_{2}e^{-t}