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

The solution for y^{''}-y^'=y 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

y^{''}-y^'=y

y^{^{''}} - y^{^{'}} = y

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

Solution steps

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

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}}

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