What is the solution for y^{''}+4y=0,y^1(0)=sin^2(x) ?

The solution for y^{''}+4y=0,y^1(0)=sin^2(x) is y=c_{1}cos(2x)+c_{2}sin(2x)

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y^{''}+4y=0,y^1(0)=sin^2(x)

y^{^{''}} + 4 y = 0, y^{1} (0) = sin^{2} (x)

y = c_{1} cos (2 x) + c_{2} sin (2 x)

Solution steps

y^{''} (x) + 4 y = 0

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x)

y = c_{1} cos (2 x) + c_{2} sin (2 x)

2 y^{^{''}} - 72 y = 0

y^{^{''}} - 3 y = 0, y (0) = 1, y^{^{'}} (0) = 5

u^{^{''}} \frac{( x )}{x} + b^{2} H (x) \frac{( x )}{x} = 0

(d^{2} - 6 d + 9) y = 0

x^{^{'''}} - 2 x^{^{''}} = 0

What is the solution for y^{''}+4y=0,y^1(0)=sin^2(x) ?
The solution for y^{''}+4y=0,y^1(0)=sin^2(x) is y=c_{1}cos(2x)+c_{2}sin(2x)