What is the solution for y^{''}+4y=1,y(0)=0,y^'(0)=1 ?

The solution for y^{''}+4y=1,y(0)=0,y^'(0)=1 is y=-1/4 cos(2t)+1/2 sin(2t)+1/4

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y^{''}+4y=1,y(0)=0,y^'(0)=1

y^{^{''}} + 4 y = 1, y (0) = 0, y^{^{'}} (0) = 1

y = - \frac{1}{4} cos (2 t) + \frac{1}{2} sin (2 t) + \frac{1}{4}

Solution steps

y^{''} (t) + 4 y = 1

Solve linear ODE:

y = - \frac{1}{4} cos (2 t) + \frac{1}{2} sin (2 t) + \frac{1}{4}

y = - \frac{1}{4} cos (2 t) + \frac{1}{2} sin (2 t) + \frac{1}{4}

What is the solution for y^{''}+4y=1,y(0)=0,y^'(0)=1 ?
The solution for y^{''}+4y=1,y(0)=0,y^'(0)=1 is y=-1/4 cos(2t)+1/2 sin(2t)+1/4