What is the solution for y^{''}-6y^'-7y=0,y(0)=b,y^'(0)=8 ?

The solution for y^{''}-6y^'-7y=0,y(0)=b,y^'(0)=8 is y=e^{7t}(b+(8-7b)/8)-(e^{-t}(8-7b))/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-6y^'-7y=0,y(0)=b,y^'(0)=8

y^{^{''}} - 6 y^{^{'}} - 7 y = 0, y (0) = b, y^{^{'}} (0) = 8

y = e^{7 t} (b + \frac{8 - 7 b}{8}) - \frac{e ^{- t} ( 8 - 7 b )}{8}

Solution steps

y^{''} (t) - 6 y^{'} (t) - 7 y = 0

Solve linear ODE:

y = e^{7 t} (b + \frac{8 - 7 b}{8}) - \frac{e ^{- t} ( 8 - 7 b )}{8}

y = e^{7 t} (b + \frac{8 - 7 b}{8}) - \frac{e ^{- t} ( 8 - 7 b )}{8}

cot^{2} (sin (x))

\int ln (x) x^{'} d x

\frac{d}{d x} (3 e^{x^{2} + x})

n = 1 \sum \infty \frac{e ^{n}}{π ^{n}}

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

What is the solution for y^{''}-6y^'-7y=0,y(0)=b,y^'(0)=8 ?
The solution for y^{''}-6y^'-7y=0,y(0)=b,y^'(0)=8 is y=e^{7t}(b+(8-7b)/8)-(e^{-t}(8-7b))/8