What is the solution for y^{''}+4y=0,y(0)=5,y^'(0)=-3 ?

The solution for y^{''}+4y=0,y(0)=5,y^'(0)=-3 is y=5cos(2t)-3/2 sin(2t)

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y^{''}+4y=0,y(0)=5,y^'(0)=-3

y^{^{''}} + 4 y = 0, y (0) = 5, y^{^{'}} (0) = - 3

y = 5 cos (2 t) - \frac{3}{2} sin (2 t)

Solution steps

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

Solve linear ODE:

y = 5 cos (2 t) - \frac{3}{2} sin (2 t)

y = 5 cos (2 t) - \frac{3}{2} sin (2 t)

(D^{2} + 64)^{2} y = 0

y^{^{''}} = a^{2} y

4 \cdot \frac{d ^{2} y}{d x ^{2}} + 12 \cdot \frac{d y}{d x} + 9 y = 0

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (0) = \frac{2}{3}

0.2 q^{^{''}} + 4 q^{^{'}} + 100 q = 0

What is the solution for y^{''}+4y=0,y(0)=5,y^'(0)=-3 ?
The solution for y^{''}+4y=0,y(0)=5,y^'(0)=-3 is y=5cos(2t)-3/2 sin(2t)