What is the solution for 3/4*x^{''}+36x=0,x^'(0)=0,x(0)=-6 ?

The solution for 3/4*x^{''}+36x=0,x^'(0)=0,x(0)=-6 is x=-6cos(4sqrt(3)t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3/4*x^{''}+36x=0,x^'(0)=0,x(0)=-6

\frac{3}{4} \cdot x^{^{''}} + 36 x = 0, x^{^{'}} (0) = 0, x (0) = - 6

x = - 6 cos (43 t)

Solution steps

\frac{3}{4} \cdot x^{''} (t) + 36 x = 0

Solve linear ODE:

x = - 6 cos (43 t)

x = - 6 cos (43 t)

y^{^{''}} + 6 y^{^{'}} + 8 y = 0, y (0) = 0, y^{^{'}} (0) = - 5

y^{^{''}} - y^{^{'}} - 72 y^{^{'}} = 0

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

\frac{y ^{^{''}}}{8} + 30 y^{^{'}} + 500 y = 0

y^{^{''}} - 36 \cdot y = 0

What is the solution for 3/4*x^{''}+36x=0,x^'(0)=0,x(0)=-6 ?
The solution for 3/4*x^{''}+36x=0,x^'(0)=0,x(0)=-6 is x=-6cos(4sqrt(3)t)