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

The solution for y^{''}+a^2y=0,y(0)=0,y^'(0)=a is y=sin(at)

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

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

y = sin (a t)

Solution steps

y^{''} (t) + a^{2} y = 0

Solve linear ODE:

y = sin (a t)

y = sin (a t)

y^{^{'''}} - 7 y^{^{''}} + y^{^{'}} + 12 y = 0

y^{^{''}} - \frac{1}{4} y = 0, y (0) = 4, y^{^{'}} (0) = 0

y^{^{''}} - 2 y^{^{'}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = - 2

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y (0) = 5, y^{^{'}} = 2

y^{^{''}} + 16 y^{^{'}} + 64 y = 0, y (0) = 5, y^{^{'}} (0) = 0

What is the solution for y^{''}+a^2y=0,y(0)=0,y^'(0)=a ?
The solution for y^{''}+a^2y=0,y(0)=0,y^'(0)=a is y=sin(at)