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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

81 y^{^{''}} + 108 y^{^{'}} - 13 y = 0, y (0) = 1, y^{^{'}} (0) = 0

x^{^{''}} + 0.5 x^{^{'}} + 4 x = 0

x^{^{''}} - c^{2} x = 0

y^{^{''}} - 8 y^{^{'}} + 65 y = 0, y (0) = 1, y^{^{'}} (0) = 2

y^{^{''}} - 9 y^{^{'}} + 22 y = 0

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