What is the solution for 3y^{''}-8y^'-3y=0,y(0)=-9,y^'(0)=8 ?

The solution for 3y^{''}-8y^'-3y=0,y(0)=-9,y^'(0)=8 is y=(-9+21/2)e^{3t}-21/2 e^{-t/3}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3y^{''}-8y^'-3y=0,y(0)=-9,y^'(0)=8

3 y^{^{''}} - 8 y^{^{'}} - 3 y = 0, y (0) = - 9, y^{^{'}} (0) = 8

y = (- 9 + \frac{21}{2}) e^{3 t} - \frac{21}{2} e^{- \frac{t}{3}}

Solution steps

3 y^{''} (t) - 8 y^{'} (t) - 3 y = 0

Solve linear ODE:

y = (- 9 + \frac{21}{2}) e^{3 t} - \frac{21}{2} e^{- \frac{t}{3}}

y = (- 9 + \frac{21}{2}) e^{3 t} - \frac{21}{2} e^{- \frac{t}{3}}

\int_{1}^{3} (\frac{3 x ^{2} - 7}{x ^{3}}) d x

\int t^{3} (7 + t^{4})^{3} d t

x \to 1 lim ((x)^{\frac{1}{x - 1}})

\int_{0}^{4} 3 x d x

\int x^{2} 36 - x^{2} d x

What is the solution for 3y^{''}-8y^'-3y=0,y(0)=-9,y^'(0)=8 ?
The solution for 3y^{''}-8y^'-3y=0,y(0)=-9,y^'(0)=8 is y=(-9+21/2)e^{3t}-21/2 e^{-t/3}