What is the solution for y^{''}+y^'-6y=0,y(0)=10,y^'(0)=0 ?

The solution for y^{''}+y^'-6y=0,y(0)=10,y^'(0)=0 is y=6e^{2t}+4e^{-3t}

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)=10,y^'(0)=0

y^{^{''}} + y^{^{'}} - 6 y = 0, y (0) = 10, y^{^{'}} (0) = 0

y = 6 e^{2 t} + 4 e^{- 3 t}

Solution steps

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

Solve linear ODE:

y = 6 e^{2 t} + 4 e^{- 3 t}

y = 6 e^{2 t} + 4 e^{- 3 t}

\int_{0}^{\frac{π}{4}} sin (2 x) d x

\frac{\partial}{\partial x} (\frac{( 2 x - y )}{x y})

\frac{d}{d x} (a ln (x))

f (x) 0

\frac{- 3 x ^{4} + 1}{( x ^{4} + 1 ) ^{2}}

What is the solution for y^{''}+y^'-6y=0,y(0)=10,y^'(0)=0 ?
The solution for y^{''}+y^'-6y=0,y(0)=10,y^'(0)=0 is y=6e^{2t}+4e^{-3t}