What is the solution for x^{''}+x=0,x(pi/2)=0,x^'(pi/2)=1 ?

The solution for x^{''}+x=0,x(pi/2)=0,x^'(pi/2)=1 is x=-cos(t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{''}+x=0,x(pi/2)=0,x^'(pi/2)=1

x^{^{''}} + x = 0, x (\frac{π}{2}) = 0, x^{^{'}} (\frac{π}{2}) = 1

x = - cos (t)

Solution steps

x^{''} (t) + x = 0

Solve linear ODE:

x = - cos (t)

x = - cos (t)

\int a^{x} x d x

y = 2 x^{2}, y = 8 x - 2 x^{2}

\frac{\partial}{\partial y} (x^{2} + y^{2} - 13)

f (x) = - x^{2} - 7 x + 44, at x = 2

\int - e^{3 x} d x

What is the solution for x^{''}+x=0,x(pi/2)=0,x^'(pi/2)=1 ?
The solution for x^{''}+x=0,x(pi/2)=0,x^'(pi/2)=1 is x=-cos(t)