What is the solution for y^{''}+4y^'-2y=0 ?

The solution for y^{''}+4y^'-2y=0 is y=c_{1}e^{(-2+sqrt(6))t}+c_{2}e^{(-2-sqrt(6))t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y^'-2y=0

y^{^{''}} + 4 y^{^{'}} - 2 y = 0

y = c_{1} e^{(- 2 + 6) t} + c_{2} e^{(- 2 - 6) t}

Solution steps

y^{''} (t) + 4 y^{'} (t) - 2 y = 0

Solve linear ODE:

y = c_{1} e^{(- 2 + 6) t} + c_{2} e^{(- 2 - 6) t}

y = c_{1} e^{(- 2 + 6) t} + c_{2} e^{(- 2 - 6) t}

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} - 6 y = 0

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

y^{^{''}} + 4 y^{^{'}} + 6 y = 0

\frac{d ^{4} y}{d x ^{4}} - 23 \frac{d ^{2} y}{d x ^{2}} - 50 y = 0

4 y^{^{''}} - 4 y^{^{'}} - 3 y = 0, y (0) = 1, y^{^{'}} (0) = 5

What is the solution for y^{''}+4y^'-2y=0 ?
The solution for y^{''}+4y^'-2y=0 is y=c_{1}e^{(-2+sqrt(6))t}+c_{2}e^{(-2-sqrt(6))t}