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

The solution for y^{''}+y^'-y=0 is y=c_{1}e^{((-1+sqrt(5))t)/2}+c_{2}e^{((-1-sqrt(5))t)/2}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = c_{1} e^{\frac{( - 1 + 5 ) t}{2}} + c_{2} e^{\frac{( - 1 - 5 ) t}{2}}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{( - 1 + 5 ) t}{2}} + c_{2} e^{\frac{( - 1 - 5 ) t}{2}}

y = c_{1} e^{\frac{( - 1 + 5 ) t}{2}} + c_{2} e^{\frac{( - 1 - 5 ) t}{2}}

\frac{\partial}{\partial x} (\frac{4 x - 4 y}{4 x + 4 y})

\int_{- 2}^{6} (4 x) - (x^{2} - 12) d x

(y^{2} + y \cdot x) d x - x^{2} d y = 0

\frac{\partial}{\partial x} (\frac{x ^{3} \cdot y}{x ^{2} + y ^{2}})

2^{n}

What is the solution for y^{''}+y^'-y=0 ?
The solution for y^{''}+y^'-y=0 is y=c_{1}e^{((-1+sqrt(5))t)/2}+c_{2}e^{((-1-sqrt(5))t)/2}