What is the solution for y^{''}+25y=0,y(0)=3,y^'(0)=1 ?

The solution for y^{''}+25y=0,y(0)=3,y^'(0)=1 is y=3cos(5t)+1/5 sin(5t)

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y^{''}+25y=0,y(0)=3,y^'(0)=1

y^{^{''}} + 25 y = 0, y (0) = 3, y^{^{'}} (0) = 1

y = 3 cos (5 t) + \frac{1}{5} sin (5 t)

Solution steps

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

Solve linear ODE:

y = 3 cos (5 t) + \frac{1}{5} sin (5 t)

y = 3 cos (5 t) + \frac{1}{5} sin (5 t)

y^{^{''}} + 25 y = 0, y (0) = 3, y^{^{'}} (0) = 2

y^{^{'''}} + 8 y^{^{''}} + 21 y^{^{'}} + 18 y = 0

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

22 y^{^{''''}} + 61 y^{^{''}} + 25 y = 0

y^{^{''}} + 5 y^{^{'}} + 6 y = 0, y (0) = 2, y^{^{'}} (0) = 5

What is the solution for y^{''}+25y=0,y(0)=3,y^'(0)=1 ?
The solution for y^{''}+25y=0,y(0)=3,y^'(0)=1 is y=3cos(5t)+1/5 sin(5t)