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

Popular >

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}}

y^{^{''}} + 4 y = 0, y (0) = 3, y^{^{'}} (0) = - 2

y^{^{''}} - 9 y^{^{'}} + 20 y = 0, y (0) = 1, y^{^{'}} (0) = 2

u^{^{''}} + 2 u^{^{'}} + 2 H (t) = 0

55 x^{^{''}} + 220 x = 0, x (0) = 0, x^{^{'}} (0) = - 6

y^{^{''}} + 2 y^{^{'}} - 8 y = 0, y (0) = 3, y^{^{'}} (0) = 12

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}