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

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

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)=-6

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

y = \frac{e ^{3 t} ( - 6 + 2 c _{2} )}{3} + c_{2} e^{- 2 t}

Solution steps

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

Solve linear ODE:

y = \frac{e ^{3 t} ( - 6 + 2 c _{2} )}{3} + c_{2} e^{- 2 t}

y = \frac{e ^{3 t} ( - 6 + 2 c _{2} )}{3} + c_{2} e^{- 2 t}

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

\int 6 e^{- 0.8 x} d x

n = 1 \sum \infty (\frac{π}{4})^{n}

\frac{d}{d x} (4 x^{2} - 5 x + 3)

\int x - x^{2} d x

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