What is the solution for y^{''}+4y=0,y(0)=0,y^'(0)= 18/5 ?

The solution for y^{''}+4y=0,y(0)=0,y^'(0)= 18/5 is y= 9/5 sin(2t)

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y^{''}+4y=0,y(0)=0,y^'(0)= 18/5

y^{^{''}} + 4 y = 0, y (0) = 0, y^{^{'}} (0) = \frac{18}{5}

y = \frac{9}{5} sin (2 t)

Solution steps

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

Solve linear ODE:

y = \frac{9}{5} sin (2 t)

y = \frac{9}{5} sin (2 t)

64 y^{^{''}} + 16 y^{^{'}} + 50 y = 0

x^{^{''}} - (nπ)^{2} x = 0

y^{^{''}} - 3 y^{^{'}} - 4 y = 0

3 y^{^{''}} + 2 y^{^{'}} - y = 0, y (0) = 1, y^{^{'}} (0) = - 3

\frac{d ^{2} y}{d t ^{2}} - 4 y = 0

What is the solution for y^{''}+4y=0,y(0)=0,y^'(0)= 18/5 ?
The solution for y^{''}+4y=0,y(0)=0,y^'(0)= 18/5 is y= 9/5 sin(2t)