What is the solution for y^{''}+4y=0,y(0)=3,y^'(0)=-2 ?

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

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y^{''}+4y=0,y(0)=3,y^'(0)=-2

y^{^{''}} + 4 y = 0, y (0) = 3, y^{^{'}} (0) = - 2

y = 3 cos (2 t) - sin (2 t)

Solution steps

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

Solve linear ODE:

y = 3 cos (2 t) - sin (2 t)

y = 3 cos (2 t) - sin (2 t)

y^{^{''}} - 9 y^{^{'}} + 20 y = 0, y (0) = 1, y^{^{'}} (0) = 2

u^{^{''}} + 2 u^{^{'}} + 2 H (t) = 0

55 x^{^{''}} + 220 x = 0, x (0) = 0, x^{^{'}} (0) = - 6

y^{^{'''}} + 3.1 y^{^{''}} + 3.1 y^{^{'}} + y = 0

y^{^{''}} + 6 y^{^{'}} + 8 y = 0, y (0) = \frac{2}{5}, y^{^{'}} (0) = \frac{2}{5}

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