What is the solution for y^{''}+4y=0,y(0)=5,y^'(0)=7 ?

The solution for y^{''}+4y=0,y(0)=5,y^'(0)=7 is y=5cos(2t)+7/2 sin(2t)

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y^{''}+4y=0,y(0)=5,y^'(0)=7

y^{^{''}} + 4 y = 0, y (0) = 5, y^{^{'}} (0) = 7

y = 5 cos (2 t) + \frac{7}{2} sin (2 t)

Solution steps

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

Solve linear ODE:

y = 5 cos (2 t) + \frac{7}{2} sin (2 t)

y = 5 cos (2 t) + \frac{7}{2} sin (2 t)

y^{^{''}} + 4 y = 0, y (0) = 5, y^{^{'}} (0) = 6

x^{^{''}} - x^{^{'}} - 12 x = 0

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

\frac{1}{8} y^{^{''}} + 4 y^{^{'}} + 12 y = 0

y^{^{''}} - 6 y^{^{'}} + 2 y = 0, y (0) = 1, y^{^{'}} (0) = 1

What is the solution for y^{''}+4y=0,y(0)=5,y^'(0)=7 ?
The solution for y^{''}+4y=0,y(0)=5,y^'(0)=7 is y=5cos(2t)+7/2 sin(2t)