What is the solution for y^{''}+36y=0,y(pi/6)=1,y^'(pi/6)=-2 ?

The solution for y^{''}+36y=0,y(pi/6)=1,y^'(pi/6)=-2 is y=-cos(6t)+1/3 sin(6t)

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y^{''}+36y=0,y(pi/6)=1,y^'(pi/6)=-2

y^{^{''}} + 36 y = 0, y (\frac{π}{6}) = 1, y^{^{'}} (\frac{π}{6}) = - 2

y = - cos (6 t) + \frac{1}{3} sin (6 t)

Solution steps

y^{''} (t) + 36 y = 0

Solve linear ODE:

y = - cos (6 t) + \frac{1}{3} sin (6 t)

y = - cos (6 t) + \frac{1}{3} sin (6 t)

y^{^{'''}} - y^{^{''}} - 24 y^{^{'}} - 36 y = 0

\frac{d ^{2} y}{d x ^{2}} + 225 y = 0

y^{^{'''}} - 27 y = 0, y (0) = - 1, y^{^{'}} (0) = 6, y^{^{'''}} (0) = 18

y^{^{'''}} - 8 y = 0, y (0) = 1, y^{^{'}} (0) = - 1, y^{^{''}} (0) = 0

x^{^{''}} + 42.1788 x^{^{'}} + 295.2064 x = 0

What is the solution for y^{''}+36y=0,y(pi/6)=1,y^'(pi/6)=-2 ?
The solution for y^{''}+36y=0,y(pi/6)=1,y^'(pi/6)=-2 is y=-cos(6t)+1/3 sin(6t)