What is the solution for y^{''}+y=u_{3}(t),y(0)=0,y^'(0)=1 ?

The solution for y^{''}+y=u_{3}(t),y(0)=0,y^'(0)=1 is y=u_{3}t+(1-u_{3})sin(t)

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y^{''}+y=u_{3}(t),y(0)=0,y^'(0)=1

y^{^{''}} + y = u_{3} (t), y (0) = 0, y^{^{'}} (0) = 1

y = u_{3} t + (1 - u_{3}) sin (t)

Solution steps

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

Solve linear ODE:

y = u_{3} t + (1 - u_{3}) sin (t)

y = u_{3} t + (1 - u_{3}) sin (t)

y^{^{''}} - 4 y^{^{'}} + 4 y = x e^{x}

y^{^{''}} + 3 y^{^{'}} + 2 y = 2 x

y^{^{''}} + 6 y^{^{'}} + 5 y = 5 x^{3} + 13 x^{2} + 4 x

y^{^{''''}} + 8 y^{^{''}} + 16 y = cos (3 x) + 5

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

What is the solution for y^{''}+y=u_{3}(t),y(0)=0,y^'(0)=1 ?
The solution for y^{''}+y=u_{3}(t),y(0)=0,y^'(0)=1 is y=u_{3}t+(1-u_{3})sin(t)