What is the solution for y^{''}-2y^'-15y=0 ?

The solution for y^{''}-2y^'-15y=0 is y=c_{1}e^{5t}+c_{2}e^{-3t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-2y^'-15y=0

y^{^{''}} - 2 y^{^{'}} - 15 y = 0

y = c_{1} e^{5 t} + c_{2} e^{- 3 t}

Solution steps

y^{''} (t) - 2 y^{'} (t) - 15 y = 0

Solve linear ODE:

y = c_{1} e^{5 t} + c_{2} e^{- 3 t}

y = c_{1} e^{5 t} + c_{2} e^{- 3 t}

(4 y + y x^{2}) d y = (2 x + x y^{2}) d x

\frac{d}{d x} (8 e^{2 x} x + 12 e^{2 x})

6 - x^{2}, 8 - 3 x

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

\int_{- 1}^{1} 2 e^{x} d x

What is the solution for y^{''}-2y^'-15y=0 ?
The solution for y^{''}-2y^'-15y=0 is y=c_{1}e^{5t}+c_{2}e^{-3t}