What is the solution for t^2y^{''}+4ty^'+2y=0 ?

The solution for t^2y^{''}+4ty^'+2y=0 is y=(c_{1})/t+(c_{2})/(t^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

t^2y^{''}+4ty^'+2y=0

t^{2} y^{^{''}} + 4 t y^{^{'}} + 2 y = 0

y = \frac{c _{1}}{t} + \frac{c _{2}}{t ^{2}}

Solution steps

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

Solve linear ODE:

y = \frac{c _{1}}{t} + \frac{c _{2}}{t ^{2}}

y = \frac{c _{1}}{t} + \frac{c _{2}}{t ^{2}}

x \to 0 + lim ((x^{2}) ln (x))

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

y = x - 1

\int \frac{8 x}{x ^{3} + x ^{2} - x - 1} d x

\frac{d}{d x} (- cos (2 x) \cdot 2)

What is the solution for t^2y^{''}+4ty^'+2y=0 ?
The solution for t^2y^{''}+4ty^'+2y=0 is y=(c_{1})/t+(c_{2})/(t^2)