What is the solution for y^{''}+4y=0,y(0)=0,y^'(0)=3 ?

The solution for y^{''}+4y=0,y(0)=0,y^'(0)=3 is y= 3/2 sin(2t)

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y^{''}+4y=0,y(0)=0,y^'(0)=3

y^{^{''}} + 4 y = 0, y (0) = 0, y^{^{'}} (0) = 3

y = \frac{3}{2} sin (2 t)

Solution steps

y^{''} (t) + 4 y = 0

Solve linear ODE:

y = \frac{3}{2} sin (2 t)

y = \frac{3}{2} sin (2 t)

y^{^{''}} + 2 y^{^{'}} + 7 y = 0

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

y^{^{''}} - 14 y^{^{'}} + 53 y = 0, y (0) = 6, y^{^{'}} (0) = - 2

2 x^{^{''}} - 2 x = 0

y^{^{''}} - 7 y^{^{'}} - 10 y = 0

What is the solution for y^{''}+4y=0,y(0)=0,y^'(0)=3 ?
The solution for y^{''}+4y=0,y(0)=0,y^'(0)=3 is y= 3/2 sin(2t)