What is the solution for x^{''}+25x=0,x(0)=0.6,x^'(0)=1 ?

The solution for x^{''}+25x=0,x(0)=0.6,x^'(0)=1 is x=0.6cos(5t)+1/5 sin(5t)

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x^{''}+25x=0,x(0)=0.6,x^'(0)=1

x^{^{''}} + 25 x = 0, x (0) = 0.6, x^{^{'}} (0) = 1

x = 0.6 cos (5 t) + \frac{1}{5} sin (5 t)

Solution steps

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

Solve linear ODE:

x = 0.6 cos (5 t) + \frac{1}{5} sin (5 t)

x = 0.6 cos (5 t) + \frac{1}{5} sin (5 t)

- a \cdot y^{^{''}} + y^{^{'}} = 0, y (0) = 1, y (1) = 0

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

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

49 y^{^{''}} + 42 y^{^{'}} + 9 y = 0, y (0) = 7, y^{^{'}} (0) = 1

y^{^{''''}} - 6 y^{^{'''}} + 5 y^{^{''}} = 0

What is the solution for x^{''}+25x=0,x(0)=0.6,x^'(0)=1 ?
The solution for x^{''}+25x=0,x(0)=0.6,x^'(0)=1 is x=0.6cos(5t)+1/5 sin(5t)