What is the solution for y^{''}+25y=3,y(0)=1,y(0)=5 ?

The solution for y^{''}+25y=3,y(0)=1,y(0)=5 is y= 22/25 cos(5t)+c_{2}sin(5t)+3/25

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y^{''}+25y=3,y(0)=1,y(0)=5

y^{^{''}} + 25 y = 3, y (0) = 1, y (0) = 5

y = \frac{22}{25} cos (5 t) + c_{2} sin (5 t) + \frac{3}{25}

Solution steps

y^{''} (t) + 25 y = 3

Solve linear ODE:

y = \frac{22}{25} cos (5 t) + c_{2} sin (5 t) + \frac{3}{25}

y = \frac{22}{25} cos (5 t) + c_{2} sin (5 t) + \frac{3}{25}

y^{^{''}} + 6 y^{^{'}} + 8 y = 2 t - 3 cos (3 t)

x^{^{''}} + 10 x^{^{'}} + 25 x = 5 sin (3 t)

y^{^{''}} - 4 y = cos (2 t)

y^{^{'''}} - y^{^{''}} = t^{2} + e^{t}

y^{^{''}} - 2 y^{^{'}} - 15 y = 480 e^{- t}

What is the solution for y^{''}+25y=3,y(0)=1,y(0)=5 ?
The solution for y^{''}+25y=3,y(0)=1,y(0)=5 is y= 22/25 cos(5t)+c_{2}sin(5t)+3/25