What is u^{'''}+3u^{''}-2\H(t)=0 ?

The answer to u^{'''}+3u^{''}-2\H(t)=0 is 0+3*0-2\H(t(x))=0

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u^{'''}+3u^{''}-2\H(t)=0

u^{'''} + 3 u^{''} - 2 H (t (x)) = 0

0 + 3 \cdot 0 - 2 H (t (x)) = 0

Solution steps

u^{'''} + 3 u^{''} - 2 H (t (x)) = 0

u^{'''} = 0

0 + 3 u^{^{''}} - 2 H (t (x)) = 0

u^{''} = 0

0 + 3 \cdot 0 - 2 H (t (x)) = 0

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 25 y = 0

y^{^{''}} + 7 y^{^{'}} - 10 y = 0

64 y^{^{''}} + 48 y^{^{'}} + 9 y = 0

x^{^{''}} + 2 x^{^{'}} + 2 x = 0, x (0) = 1, x^{^{'}} (0) = 0

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

What is u^{'''}+3u^{''}-2\H(t)=0 ?
The answer to u^{'''}+3u^{''}-2\H(t)=0 is 0+3*0-2\H(t(x))=0