What is 3u^{''}+38u^'+144\H(t)=0 ?

The answer to 3u^{''}+38u^'+144\H(t)=0 is 3*0+38*0+144\H(t(x))=0

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3u^{''}+38u^'+144\H(t)=0

3 u^{''} + 38 u^{'} + 144 H (t (x)) = 0

3 \cdot 0 + 38 \cdot 0 + 144 H (t (x)) = 0

Solution steps

3 u^{''} + 38 u^{'} + 144 H (t (x)) = 0

u^{''} = 0

3 \cdot 0 + 38 u^{^{'}} + 144 H (t (x)) = 0

u^{'} = 0

3 \cdot 0 + 38 \cdot 0 + 144 H (t (x)) = 0

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (0) = - 1, y^{^{'}} (0) = - \frac{7}{2}

y^{^{''''}} + 18 y^{^{''}} - 81 y = 0

y^{^{''}} + 10 y^{^{'}} + 288 y = 0

(d^{2} + 6) y = 0

y^{^{''}} + 9.82 \cdot y^{^{'}} = 0

What is 3u^{''}+38u^'+144\H(t)=0 ?
The answer to 3u^{''}+38u^'+144\H(t)=0 is 3*0+38*0+144\H(t(x))=0