What is the solution for y^{''}+2y^'+y=0,y^'(0)=6,y(1)=6 ?

The solution for y^{''}+2y^'+y=0,y^'(0)=6,y(1)=6 is y=(-6+(6e+6)/2)e^{-t}+((6e+6)e^{-t}t)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+2y^'+y=0,y^'(0)=6,y(1)=6

y^{^{''}} + 2 y^{^{'}} + y = 0, y^{^{'}} (0) = 6, y (1) = 6

y = (- 6 + \frac{6 e + 6}{2}) e^{- t} + \frac{( 6 e + 6 ) e ^{- t} t}{2}

Solution steps

y^{''} (t) + 2 y^{'} (t) + y = 0

Solve linear ODE:

y = (- 6 + \frac{6 e + 6}{2}) e^{- t} + \frac{( 6 e + 6 ) e ^{- t} t}{2}

y = (- 6 + \frac{6 e + 6}{2}) e^{- t} + \frac{( 6 e + 6 ) e ^{- t} t}{2}

y^{^{'''}} - 10 y^{^{''}} + 9 y^{^{'}} = 0

y^{^{''}} + 64 y = 0, y (0) = 8, y^{^{'}} (0) = - \frac{4}{3}

\frac{d ^{2} x}{d t ^{2}} + 324 \frac{d x}{d t} + 4 x = 0

x^{^{''}} + 5 \cdot x = 0

y^{^{''}} + 121 y = 0

What is the solution for y^{''}+2y^'+y=0,y^'(0)=6,y(1)=6 ?
The solution for y^{''}+2y^'+y=0,y^'(0)=6,y(1)=6 is y=(-6+(6e+6)/2)e^{-t}+((6e+6)e^{-t}t)/2