What is the solution for y^{''}+4y=0,y(0)=0,y^'(0)= 12/5 ?

The solution for y^{''}+4y=0,y(0)=0,y^'(0)= 12/5 is y= 6/5 sin(2t)

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y^{''}+4y=0,y(0)=0,y^'(0)= 12/5

y^{^{''}} + 4 y = 0, y (0) = 0, y^{^{'}} (0) = \frac{12}{5}

y = \frac{6}{5} sin (2 t)

Solution steps

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

Solve linear ODE:

y = \frac{6}{5} sin (2 t)

y = \frac{6}{5} sin (2 t)

What is the solution for y^{''}+4y=0,y(0)=0,y^'(0)= 12/5 ?
The solution for y^{''}+4y=0,y(0)=0,y^'(0)= 12/5 is y= 6/5 sin(2t)