What is the solution for y^{''}-36y=8,y(0)=1,y^'(0)=0 ?

The solution for y^{''}-36y=8,y(0)=1,y^'(0)=0 is y=(-11/18+11/9)e^{6t}+11/18 e^{-6t}-2/9

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}-36y=8,y(0)=1,y^'(0)=0

y^{^{''}} - 36 y = 8, y (0) = 1, y^{^{'}} (0) = 0

y = (- \frac{11}{18} + \frac{11}{9}) e^{6 t} + \frac{11}{18} e^{- 6 t} - \frac{2}{9}

Solution steps

y^{''} (t) - 36 y = 8

Solve linear ODE:

y = (- \frac{11}{18} + \frac{11}{9}) e^{6 t} + \frac{11}{18} e^{- 6 t} - \frac{2}{9}

y = (- \frac{11}{18} + \frac{11}{9}) e^{6 t} + \frac{11}{18} e^{- 6 t} - \frac{2}{9}

What is the solution for y^{''}-36y=8,y(0)=1,y^'(0)=0 ?
The solution for y^{''}-36y=8,y(0)=1,y^'(0)=0 is y=(-11/18+11/9)e^{6t}+11/18 e^{-6t}-2/9