What is the solution for y^{''}+y=0,y(pi/3)=0,y^'(pi/3)=4 ?

The solution for y^{''}+y=0,y(pi/3)=0,y^'(pi/3)=4 is y=-sqrt(3)*2cos(t)+2sin(t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

y^{^{''}} + y = 0, y (\frac{π}{3}) = 0, y^{^{'}} (\frac{π}{3}) = 4

y = - 3 \cdot 2 cos (t) + 2 sin (t)

Solution steps

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

Solve linear ODE:

y = - 3 \cdot 2 cos (t) + 2 sin (t)

y = - 3 \cdot 2 cos (t) + 2 sin (t)

\frac{d}{d x} ((tan (\frac{1}{x}))^{s e c (\frac{1}{x})})

\int_{6}^{10} \frac{1}{( x - 6 ) ^{4}} d x

\int \frac{x}{2 x + 4} d x

x \to 3 lim (\frac{12 - x - 3}{1 + x - x ^{2} + 7})

\frac{d}{d y} (- 2 x y sin (x^{2} y))

What is the solution for y^{''}+y=0,y(pi/3)=0,y^'(pi/3)=4 ?
The solution for y^{''}+y=0,y(pi/3)=0,y^'(pi/3)=4 is y=-sqrt(3)*2cos(t)+2sin(t)