What is the solution for y^{''}+25y=0,y(0)=4.6,y^'(0)=-1.2 ?

The solution for y^{''}+25y=0,y(0)=4.6,y^'(0)=-1.2 is y=4.6cos(5t)-0.24sin(5t)

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y^{''}+25y=0,y(0)=4.6,y^'(0)=-1.2

y^{^{''}} + 25 y = 0, y (0) = 4.6, y^{^{'}} (0) = - 1.2

y = 4.6 cos (5 t) - 0.24 sin (5 t)

Solution steps

y^{''} (t) + 25 y = 0

Solve linear ODE:

y = 4.6 cos (5 t) - 0.24 sin (5 t)

y = 4.6 cos (5 t) - 0.24 sin (5 t)

\int x^{5} cos (d) x d x

\frac{\partial}{\partial x} (e^{2 x y})

\frac{d}{d x} (4 + 8 x - 5 x^{2})

\int 8 e^{4 y} d y

\frac{1}{( s ^{2} + 3 )}

What is the solution for y^{''}+25y=0,y(0)=4.6,y^'(0)=-1.2 ?
The solution for y^{''}+25y=0,y(0)=4.6,y^'(0)=-1.2 is y=4.6cos(5t)-0.24sin(5t)