What is the solution for y^{''}+y^'-6y=0,y(0)=1,y^'(0)=k ?

The solution for y^{''}+y^'-6y=0,y(0)=1,y^'(0)=k is y=e^{2t}(1+(k-2)/5)-(e^{-3t}(k-2))/5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+y^'-6y=0,y(0)=1,y^'(0)=k

y^{^{''}} + y^{^{'}} - 6 y = 0, y (0) = 1, y^{^{'}} (0) = k

y = e^{2 t} (1 + \frac{k - 2}{5}) - \frac{e ^{- 3 t} ( k - 2 )}{5}

Solution steps

y^{''} (t) + y^{'} (t) - 6 y = 0

Solve linear ODE:

y = e^{2 t} (1 + \frac{k - 2}{5}) - \frac{e ^{- 3 t} ( k - 2 )}{5}

y = e^{2 t} (1 + \frac{k - 2}{5}) - \frac{e ^{- 3 t} ( k - 2 )}{5}

(1 + \frac{3}{y} + x) (\frac{d y}{d x}) + y = \frac{3}{x} - 1

(9, - 10), (- 10, 3)

\frac{\partial}{\partial x} (x^{5} + 4^{y} + x^{y})

x \to \infty lim (4)

\int_{0}^{9} 2 - x d x

What is the solution for y^{''}+y^'-6y=0,y(0)=1,y^'(0)=k ?
The solution for y^{''}+y^'-6y=0,y(0)=1,y^'(0)=k is y=e^{2t}(1+(k-2)/5)-(e^{-3t}(k-2))/5