What is the solution for 2y^{''}-7y^'+3y=0,y(0)=0,y^'(0)=2 ?

The solution for 2y^{''}-7y^'+3y=0,y(0)=0,y^'(0)=2 is y= 4/5 e^{3t}-4/5 e^{t/2}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2y^{''}-7y^'+3y=0,y(0)=0,y^'(0)=2

2 y^{^{''}} - 7 y^{^{'}} + 3 y = 0, y (0) = 0, y^{^{'}} (0) = 2

y = \frac{4}{5} e^{3 t} - \frac{4}{5} e^{\frac{t}{2}}

Solution steps

2 y^{''} (t) - 7 y^{'} (t) + 3 y = 0

Solve linear ODE:

y = \frac{4}{5} e^{3 t} - \frac{4}{5} e^{\frac{t}{2}}

y = \frac{4}{5} e^{3 t} - \frac{4}{5} e^{\frac{t}{2}}

625 y^{^{''''}} = y

y^{^{''}} + 14 y^{^{'}} = 0, y (0) = - 2, y^{^{'}} (0) = 14

y^{^{''}} + y^{^{'}} - 2 y = 0, y (0) = 4, y^{^{'}} (0) = 5

y^{^{''}} + 3 y^{^{'}} + 12 y = 0

\frac{4}{9} y^{^{''}} + 36 y = 0

What is the solution for 2y^{''}-7y^'+3y=0,y(0)=0,y^'(0)=2 ?
The solution for 2y^{''}-7y^'+3y=0,y(0)=0,y^'(0)=2 is y= 4/5 e^{3t}-4/5 e^{t/2}