What is the solution for u^{''}+\H(t)=0,\H(0)=0,u^'(0)=2 ?

The solution for u^{''}+\H(t)=0,\H(0)=0,u^'(0)=2 is u(t)=integral of integral of-\H(t)

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

u^{^{''}} + H (t) = 0, u^{^{'}} (0) = 2

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

Solution steps

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

Isolate

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

Solve

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

Solve

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

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

2 y^{^{''}} + 2 y^{^{'}} + 8 y = 0

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} = 0

y^{^{''}} + 2 y^{^{'}} + y = 0, y (0) = - 3, y^{^{'}} (0) = 1

y^{^{''}} + 2 y^{^{'}} + y = 0, y (0) = - 3, y^{^{'}} (0) = 5

y^{^{''}} - y = 0, y^{^{'}} (0) = 0

What is the solution for u^{''}+\H(t)=0,\H(0)=0,u^'(0)=2 ?
The solution for u^{''}+\H(t)=0,\H(0)=0,u^'(0)=2 is u(t)=integral of integral of-\H(t)