What is the solution for y^{''}+6y^'-7y=0,y(0)=1,y^'(0)=1 ?

The solution for y^{''}+6y^'-7y=0,y(0)=1,y^'(0)=1 is y=e^t

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)=1,y^'(0)=1

y^{^{''}} + 6 y^{^{'}} - 7 y = 0, y (0) = 1, y^{^{'}} (0) = 1

y = e^{t}

Solution steps

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

Solve linear ODE:

y = e^{t}

y = e^{t}

f (x) = \frac{5}{4 x}

\frac{d}{d x} (\frac{2 x + 9}{9 - x})

\int_{2}^{\infty} \frac{1}{4 + x ^{2}} d x

\int \frac{x ^{2} + x + 1}{( x ^{4} + 2 x ^{2} + 1 ) ^{2}} d x

\int x^{2} arctan (6 x) d x

What is the solution for y^{''}+6y^'-7y=0,y(0)=1,y^'(0)=1 ?
The solution for y^{''}+6y^'-7y=0,y(0)=1,y^'(0)=1 is y=e^t