What is the solution for u^{''}+144\H(t)=0,\H(0)=3,u^'(0)=-48 ?

The solution for u^{''}+144\H(t)=0,\H(0)=3,u^'(0)=-48 is u(t)=integral of integral of-144\H(t)

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u^{''}+144\H(t)=0,\H(0)=3,u^'(0)=-48

u^{^{''}} + 144 H (t) = 0, u^{^{'}} (0) = - 48

u (t) = \int\int - 144 H (t) d t d t

Solution steps

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

Isolate

u^{''} (t) : u^{''} (t) = - 144 H (t)

Solve

u^{''} (t) = - 144 H (t) : u^{'} (t) = \int - 144 H (t) d t

Solve

u^{'} (t) = \int - 144 H (t) d t : u (t) = \int \int - 144 H (t) d t d t

u (t) = \int \int - 144 H (t) d t d t

y^{^{'''}} - 7 y^{^{''}} + 19 y^{^{'}} - 13 y = 0

y^{^{''}} + 2 k y^{^{'}} + k^{2} y = 0

y^{^{''}} + 2 y^{^{'}} + 63 y = 0

4 x^{^{''}} + 12 x^{^{'}} + 9 x = 0

y^{^{'''}} + 10 y^{^{''}} + 9 y^{^{'}} = 0

What is the solution for u^{''}+144\H(t)=0,\H(0)=3,u^'(0)=-48 ?
The solution for u^{''}+144\H(t)=0,\H(0)=3,u^'(0)=-48 is u(t)=integral of integral of-144\H(t)