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

2 y^{^{''}} + 20 y^{^{'}} + 50 y = 10 sin (3 x)

y^{^{'''}} - y^{^{'}} = 3 x + 4 e^{x} + 5 e^{2 x}

y^{^{'''}} - 19 y^{^{'}} + 30 y = x^{2} \cdot e^{2 x}

y^{^{''}} - 4 y^{^{'}} - 5 y = 125, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 3 y^{^{'}} + 2 y = (1 + 2 x) e^{x}

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