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

The solution for x^{''}+x^'-x=0 is x=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 >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{3} y}{d x ^{3}} - y = 0

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

5 y^{^{''}} + 10 y^{^{'}} + 5 y = 0, y (0) = 0, y^{^{'}} (0) = 2

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = 4

\frac{d ^{2} y}{d t ^{2}} - 2 \frac{d y}{d t} - 3 y = 0

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