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

The solution for y^{''}-4y^'+y=0 is y=c_{1}e^{(2+sqrt(3))t}+c_{2}e^{(2-sqrt(3))t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = c_{1} e^{(2 + 3) t} + c_{2} e^{(2 - 3) t}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{(2 + 3) t} + c_{2} e^{(2 - 3) t}

y = c_{1} e^{(2 + 3) t} + c_{2} e^{(2 - 3) t}

\int \frac{1}{9 + x ^{2}} d x

\int_{0}^{2} (2^{x} + 6) d x

\frac{\partial}{\partial v} (a sin (u) cos (v))

\frac{d}{d x} (∣ x - 1 ∣ + ∣ 2 x + 3 ∣)

y = 2 x^{2}, y = 8 x - 6 x^{2}

What is the solution for y^{''}-4y^'+y=0 ?
The solution for y^{''}-4y^'+y=0 is y=c_{1}e^{(2+sqrt(3))t}+c_{2}e^{(2-sqrt(3))t}