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

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

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

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

y = sin (5 t) + 1

Solution steps

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

Solve linear ODE:

y = sin (5 t) + 1

y = sin (5 t) + 1

y^{^{''}} - 2 y^{^{'}} - 3 y = e^{4 x}

y^{^{''}} + 10 y^{^{'}} + 25 y = - 8 - 5 x + 25 x^{2}

10 x^{^{''}} + 200 x = 600 sin (125 t)

5 y^{^{''}} + 6 y^{^{'}} + y = x^{3} + 18 x^{2} + 32 x + 13

y^{^{''}} - 6 y^{^{'}} + 8 y = 3 + e^{- 2 t}

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