What is the solution for y^{''}+36y=0,y^'(0)=2 ?

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

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y^{''}+36y=0,y^'(0)=2

y^{^{''}} + 36 y = 0, y^{^{'}} (0) = 2

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{3} y}{d x ^{3}} + y = 0

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

y^{^{''}} - 4 y^{^{'}} - 3 y = 0, y (0) = 1, y^{^{'}} (0) = 5

\frac{15}{16} u^{^{''}} + 128 H (t) = 0

y^{^{'''}} - 4 y^{^{''}} + y^{^{'}} - 4 y = 0

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