What is the solution for 4y^{''}+(16)/(0.25)y=0,y(0)=0,y^'(0)=0 ?

The solution for 4y^{''}+(16)/(0.25)y=0,y(0)=0,y^'(0)=0 is y=0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4y^{''}+(16)/(0.25)y=0,y(0)=0,y^'(0)=0

4 y^{^{''}} + \frac{16}{0.25} y = 0, y (0) = 0, y^{^{'}} (0) = 0

y = 0

Solution steps

4 y^{''} (t) + \frac{16}{0.25} y = 0

Solve linear ODE:

y = 0

y = 0

\frac{16}{32} x^{^{''}} + x^{^{'}} + 2 x = 0, x (0) = - 2, x^{^{'}} (0) = 1

100 y^{^{'''}} + 60 y^{^{''}} + 9 y^{^{'}} = 0

8 y - 2 y^{^{'}} + 6 y^{^{''}} - y^{^{'}} = 0

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y^{^{'}} 0 = (3), y (0) = 0

y^{^{''}} - (a + 1) y^{^{'}} + a y = 0

What is the solution for 4y^{''}+(16)/(0.25)y=0,y(0)=0,y^'(0)=0 ?
The solution for 4y^{''}+(16)/(0.25)y=0,y(0)=0,y^'(0)=0 is y=0