What is u^{''}+0.25u^'+\H(t)=0 ?

The answer to u^{''}+0.25u^'+\H(t)=0 is 0+0.25*0+\H(t(x))=0

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u^{''}+0.25u^'+\H(t)=0

u^{''} + 0.25 u^{'} + H (t (x)) = 0

0 + 0.25 \cdot 0 + H (t (x)) = 0

Solution steps

u^{''} + 0.25 u^{'} + H (t (x)) = 0

u^{''} = 0

0 + 0.25 u^{^{'}} + H (t (x)) = 0

u^{'} = 0

0 + 0.25 \cdot 0 + H (t (x)) = 0

y^{^{'''}} - 23 y^{^{''}} + 160 y^{^{'}} - 300 y = 0

y^{^{''}} - 2 y^{^{'}} + 2 y = 0, y (π) = 7, y (0) = 7

y^{^{''}} - 4 y^{^{'}} + 5 y = 0

y^{^{''}} + 5 y^{^{'}} + 5 y = 0

y^{^{''}} + 8 y^{^{'}} + 25 y = 0, y (0) = 2, y^{^{'}} (0) = - 7

What is u^{''}+0.25u^'+\H(t)=0 ?
The answer to u^{''}+0.25u^'+\H(t)=0 is 0+0.25*0+\H(t(x))=0