What is the solution for y^{''}+25y=0,y(0)=4,y^'(0)=-2 ?

The solution for y^{''}+25y=0,y(0)=4,y^'(0)=-2 is y=4cos(5t)-2/5 sin(5t)

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y^{''}+25y=0,y(0)=4,y^'(0)=-2

y^{^{''}} + 25 y = 0, y (0) = 4, y^{^{'}} (0) = - 2

y = 4 cos (5 t) - \frac{2}{5} sin (5 t)

Solution steps

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

Solve linear ODE:

y = 4 cos (5 t) - \frac{2}{5} sin (5 t)

y = 4 cos (5 t) - \frac{2}{5} sin (5 t)

What is the solution for y^{''}+25y=0,y(0)=4,y^'(0)=-2 ?
The solution for y^{''}+25y=0,y(0)=4,y^'(0)=-2 is y=4cos(5t)-2/5 sin(5t)