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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 36 y = 0, y (\frac{π}{4}) = 1, y^{^{'}} (\frac{π}{4}) = - 2

y^{^{''}} + 3 y^{^{'}} - y = 0, y (1) = 2, y^{^{'}} (1) = 0

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (1) = 1, y^{^{'}} (1) = 4

y^{^{'''}} + 2 y^{^{''}} - 4 y^{^{'}} - 8 y = 0

12 y^{^{''}} + 5 y^{^{'}} - 2 y = 0

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