What is the solution for y^{''}+25y=0,y(0)=4,y^'(0)=-5 ?

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

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y^{''}+25y=0,y(0)=4,y^'(0)=-5

y^{^{''}} + 25 y = 0, y (0) = 4, y^{^{'}} (0) = - 5

y = 4 cos (5 t) - sin (5 t)

Solution steps

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

Solve linear ODE:

y = 4 cos (5 t) - sin (5 t)

y = 4 cos (5 t) - sin (5 t)

y^{^{''}} + 6 y + 9 y = 0

y^{^{'''}} - 6 y^{^{''}} + 12 y^{^{'}} - 8 y = 0

y^{^{''}} + 8 y^{^{'}} + 9 y = 0, y (1) = 1, y^{^{'}} (1) = 0

2 y^{^{''}} - 4 y^{^{'}} + 3 y = 0

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

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