What is the solution for y^{''}+9y=0,y(pi/3)=3,y^'(pi/3)=3 ?

The solution for y^{''}+9y=0,y(pi/3)=3,y^'(pi/3)=3 is y=-3cos(3t)-sin(3t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+9y=0,y(pi/3)=3,y^'(pi/3)=3

y^{^{''}} + 9 y = 0, y (\frac{π}{3}) = 3, y^{^{'}} (\frac{π}{3}) = 3

y = - 3 cos (3 t) - sin (3 t)

Solution steps

y^{''} (t) + 9 y = 0

Solve linear ODE:

y = - 3 cos (3 t) - sin (3 t)

y = - 3 cos (3 t) - sin (3 t)

y^{^{''}} + y^{^{'}} - 2 y = e^{- 2 x}

\int \frac{( 2 + x )}{9 + x ^{2}} d x

\int_{0}^{1} x (\frac{2 ( x + 3 )}{7}) d x

f (x) = - 3 cos (x)

\int \frac{x ^{2}}{4 x ^{3} + 5} d x

What is the solution for y^{''}+9y=0,y(pi/3)=3,y^'(pi/3)=3 ?
The solution for y^{''}+9y=0,y(pi/3)=3,y^'(pi/3)=3 is y=-3cos(3t)-sin(3t)