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)+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) + sin (5 t) + \frac{3}{25}

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 20 y^{^{'}} + 200 y = 400

y^{^{'''}} - 7 y^{^{''}} = x^{2} + 2 cos (x)

y^{^{''}} + y = tan^{2} (t)

y^{^{''''}} + 2 y^{^{''}} + y = 9 + cos (6 t)

y^{^{''}} + y^{^{'}} - 2 y = 2 + 5 cos (2 x)

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)+sin(5t)+3/25