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)

2 y^{^{''}} + 5 y^{^{'}} - 3 y = 0, y (0) = - 5, y^{^{'}} (0) = 29

2 x^{^{''}} + 3 x^{^{'}} - 2 x = 0

(d^{4} - 6 d^{3} - 12 d^{2} + 38 d - 21) y = 0

20 y^{^{''}} + 80 y^{^{'}} + 60 y = 0

\frac{d ^{2} y}{d t ^{2}} + 9 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)