What is the solution for y^{''}-2y^'-35y=0 ?

The solution for y^{''}-2y^'-35y=0 is y=c_{1}e^{7t}+c_{2}e^{-5t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-2y^'-35y=0

y^{^{''}} - 2 y^{^{'}} - 35 y = 0

y = c_{1} e^{7 t} + c_{2} e^{- 5 t}

Solution steps

y^{''} (t) - 2 y^{'} (t) - 35 y = 0

Solve linear ODE:

y = c_{1} e^{7 t} + c_{2} e^{- 5 t}

y = c_{1} e^{7 t} + c_{2} e^{- 5 t}

\int \frac{x ^{2}}{16 + x ^{6}} d x

\int \frac{( 1 + x ^{2} ) ^{\frac{1}{2}}}{x} d x

x \to + π lim (x cos (\frac{7 x}{6}))

\int 2 x yz e^{x^{2}} d x

\int \frac{4}{1 + t ^{2}} d t

What is the solution for y^{''}-2y^'-35y=0 ?
The solution for y^{''}-2y^'-35y=0 is y=c_{1}e^{7t}+c_{2}e^{-5t}