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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} y}{d t ^{2}} - 5 \frac{d y}{d t} = 0

y^{^{'''}} - 7 y^{^{''}} = 0

\frac{4}{32} x^{^{''}} + x^{^{'}} + 2 x = 0, x (0) = - 1, x^{^{'}} (0) = 16

\frac{d ^{2} y}{d t ^{2}} - 5 \frac{d y}{d t} + 6 y = 0, y (1) = 3

x^{^{''}} - 3 x^{^{'}} = 0

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