What is the solution for y^{''}+4y=0,y(0)=2,y^'(0)=-8 ?

The solution for y^{''}+4y=0,y(0)=2,y^'(0)=-8 is y=2cos(2t)-4sin(2t)

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y^{''}+4y=0,y(0)=2,y^'(0)=-8

y^{^{''}} + 4 y = 0, y (0) = 2, y^{^{'}} (0) = - 8

y = 2 cos (2 t) - 4 sin (2 t)

Solution steps

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

Solve linear ODE:

y = 2 cos (2 t) - 4 sin (2 t)

y = 2 cos (2 t) - 4 sin (2 t)

y^{^{''}} + 25 y = 0, y (0) = 0, y^{^{'}} (0) = 20

y^{^{'''}} + 10 y^{^{''}} + 12 y^{^{'}} - 72 y = 0

\frac{d ^{2} y}{d x ^{2}} - 12 \frac{d y}{d x} + 32 y = 0

y^{^{''}} + 4 y^{^{'}} + 12 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (0) = \frac{1}{2}, y^{^{'}} (0) = \frac{1}{4}

What is the solution for y^{''}+4y=0,y(0)=2,y^'(0)=-8 ?
The solution for y^{''}+4y=0,y(0)=2,y^'(0)=-8 is y=2cos(2t)-4sin(2t)