What is the solution for t^2y^{''}+19ty^'+106y=0 ?

The solution for t^2y^{''}+19ty^'+106y=0 is y=(c_{1}cos(5ln(t))+c_{2}sin(5ln(t)))/(t^9)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

t^2y^{''}+19ty^'+106y=0

t^{2} y^{^{''}} + 19 t y^{^{'}} + 106 y = 0

y = \frac{c _{1} cos ( 5 ln ( t ) ) + c _{2} sin ( 5 ln ( t ) )}{t ^{9}}

Solution steps

t^{2} y^{''} (t) + 19 t y^{'} (t) + 106 y = 0

Solve linear ODE:

y = \frac{c _{1} cos ( 5 ln ( t ) ) + c _{2} sin ( 5 ln ( t ) )}{t ^{9}}

y = \frac{c _{1} cos ( 5 ln ( t ) ) + c _{2} sin ( 5 ln ( t ) )}{t ^{9}}

x \to 0 lim (tan (\frac{1}{x}))

f (x) = x^{- 2}, at x = - 2

\frac{\partial}{\partial p} (\frac{( p + q )}{p + r})

x \to 0 lim (\frac{sin ( 4 x )}{cos ( 5 x )})

\frac{d}{d x} (- 3 x^{3} + 2 x^{2} + 5 x - 2)

What is the solution for t^2y^{''}+19ty^'+106y=0 ?
The solution for t^2y^{''}+19ty^'+106y=0 is y=(c_{1}cos(5ln(t))+c_{2}sin(5ln(t)))/(t^9)